18

我想按字典顺序对大量整数(比如 1 百万元素)进行排序。

例子:

input [] = { 100, 21 , 22 , 99 , 1  , 927 }
sorted[] = { 1  , 100, 21 , 22 , 927, 99  }

我已经使用最简单的方法完成了它:

  • 将所有数字转换为字符串(非常昂贵,因为它会占用大量内存)
  • 使用std:sortwithstrcmp作为比较函数
  • 将字符串转换回整数

还有比这更好的方法吗?

4

12 回答 12

16

std::sort()与合适的比较功能一起使用。这减少了内存需求。

比较函数可以使用n % 10,n / 10 % 10n / 100 % 10来访问单个数字(对于正整数;负整数的工作方式略有不同)。

于 2013-10-25T11:50:36.043 回答
8

要提供任何自定义排序顺序,您可以为std::sort. 在这种情况下,它会有点复杂,使用对数来检查以 10 为底的数字的各个数字。

这是一个例子——内联评论描述了正在发生的事情。

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cassert>

int main() {
    int input[] { 100, 21, 22, 99, 1, 927, -50, -24, -160 };

    /**
     * Sorts the array lexicographically.
     * 
     * The trick is that we have to compare digits left-to-right
     * (considering typical Latin decimal notation) and that each of
     * two numbers to compare may have a different number of digits.
     * 
     * This is very efficient in storage space, but inefficient in
     * execution time; an approach that pre-visits each element and
     * stores a translated representation will at least double your
     * storage requirements (possibly a problem with large inputs)
     * but require only a single translation of each element.
     */
    std::sort(
        std::begin(input),
        std::end(input),
        [](int lhs, int rhs) -> bool {
            // Returns true if lhs < rhs
            // Returns false otherwise
            const auto BASE      = 10;
            const bool LHS_FIRST = true;
            const bool RHS_FIRST = false;
            const bool EQUAL     = false;


            // There's no point in doing anything at all
            // if both inputs are the same; strict-weak
            // ordering requires that we return `false`
            // in this case.
            if (lhs == rhs) {
                return EQUAL;
            }

            // Compensate for sign
            if (lhs < 0 && rhs < 0) {
                // When both are negative, sign on its own yields
                // no clear ordering between the two arguments.
                // 
                // Remove the sign and continue as for positive
                // numbers.
                lhs *= -1;
                rhs *= -1;
            }
            else if (lhs < 0) {
                // When the LHS is negative but the RHS is not,
                // consider the LHS "first" always as we wish to
                // prioritise the leading '-'.
                return LHS_FIRST;
            }
            else if (rhs < 0) {
                // When the RHS is negative but the LHS is not,
                // consider the RHS "first" always as we wish to
                // prioritise the leading '-'.
                return RHS_FIRST;
            }

            // Counting the number of digits in both the LHS and RHS
            // arguments is *almost* trivial.
            const auto lhs_digits = (
                lhs == 0
                ? 1
                : std::ceil(std::log(lhs+1)/std::log(BASE))
            );

            const auto rhs_digits = (
                rhs == 0
                ? 1
                : std::ceil(std::log(rhs+1)/std::log(BASE))
            );

            // Now we loop through the positions, left-to-right,
            // calculating the digit at these positions for each
            // input, and comparing them numerically. The
            // lexicographic nature of the sorting comes from the
            // fact that we are doing this per-digit comparison
            // rather than considering the input value as a whole.
            const auto max_pos = std::max(lhs_digits, rhs_digits);
            for (auto pos = 0; pos < max_pos; pos++) {
                if (lhs_digits - pos == 0) {
                    // Ran out of digits on the LHS;
                    // prioritise the shorter input
                    return LHS_FIRST;
                }
                else if (rhs_digits - pos == 0) {
                    // Ran out of digits on the RHS;
                    // prioritise the shorter input
                    return RHS_FIRST;
                }
                else {
                    const auto lhs_x = (lhs / static_cast<decltype(BASE)>(std::pow(BASE, lhs_digits - 1 - pos))) % BASE;
                    const auto rhs_x = (rhs / static_cast<decltype(BASE)>(std::pow(BASE, rhs_digits - 1 - pos))) % BASE;

                    if (lhs_x < rhs_x)
                        return LHS_FIRST;
                    else if (rhs_x < lhs_x)
                        return RHS_FIRST;
                }
            }

            // If we reached the end and everything still
            // matches up, then something probably went wrong
            // as I'd have expected to catch this in the tests
            // for equality.
            assert("Unknown case encountered");
        }
    );

    std::cout << '{';
    for (auto x : input)
        std::cout << x << ", ";
    std::cout << '}';

    // Output: {-160, -24, -50, 1, 100, 21, 22, 927, 99, }
}

演示

有更快的方法来计算数字中的位数,但以上内容将帮助您入门。

于 2013-10-25T11:58:24.330 回答
6

这是一个比较解决方案的社区 wiki。我采用了 nim 的代码并使其易于扩展。随意添加您的解决方案和输出。

示例使用 g++4.9 @ 运行一台旧的慢速计算机(3 GB RAM,Core2Duo U9400)-O3 -march=native

元素数量:1e+03
整数类型的大小:4

参考解决方案:轨道中的轻盈竞赛

解决方案“dyp”:
    持续时间:0 毫秒和 301 微秒
    与参考解决方案的比较:完全匹配
解决方案“尼姆”:
    持续时间:2 毫秒和 160 微秒
    与参考解决方案的比较:完全匹配
解决方案“nyarlathotep”:
    持续时间:8 毫秒和 126 微秒
    与参考解决方案的比较:完全匹配
解决方案“不错”:
    持续时间:1 毫秒和 102 微秒
    与参考解决方案的比较:完全匹配
解决方案“Eric Postpischil”:
    持续时间:2 毫秒和 550 微秒
    与参考解决方案的比较:完全匹配
解决方案“轨道中的亮度竞赛”:
    持续时间:17 毫秒和 469 微秒
    与参考解决方案的比较:完全匹配
解决方案“pts”:
    持续时间:1 毫秒和 92 微秒
    与参考解决方案的比较:完全匹配

==================================================== ========

元素数量:1e+04
整数类型的大小:4

参考解决方案:轨道中的轻盈竞赛

解决方案“nyarlathotep”:
    持续时间:109 毫秒和 712 微秒
    与参考解决方案的比较:完全匹配
解决方案“轨道中的亮度竞赛”:
    持续时间:272 毫秒和 819 微秒
    与参考解决方案的比较:完全匹配
解决方案“dyp”:
    持续时间:1 毫秒和 748 微秒
    与参考解决方案的比较:完全匹配
解决方案“不错”:
    持续时间:16 毫秒和 115 微秒
    与参考解决方案的比较:完全匹配
解决方案“pts”:
    持续时间:15 毫秒和 10 微秒
    与参考解决方案的比较:完全匹配
解决方案“Eric Postpischil”:
    持续时间:33 毫秒和 301 微秒
    与参考解决方案的比较:完全匹配
解决方案“尼姆”:
    持续时间:17 毫秒和 83 微秒
    与参考解决方案的比较:完全匹配

==================================================== ========

元素数量:1e+05
整数类型的大小:4

参考解决方案:轨道中的轻盈竞赛

解决方案“尼姆”:
    持续时间:217 毫秒和 4 微秒
    与参考解决方案的比较:完全匹配
解决方案“pts”:
    持续时间:199 毫秒和 505 微秒
    与参考解决方案的比较:完全匹配
解决方案“dyp”:
    持续时间:20 毫秒和 330 微秒
    与参考解决方案的比较:完全匹配
解决方案“Eric Postpischil”:
    持续时间:415 毫秒和 477 微秒
    与参考解决方案的比较:完全匹配
解决方案“轨道中的亮度竞赛”:
    持续时间:3955 毫秒和 58 微秒
    与参考解决方案的比较:完全匹配
解决方案“不错”:
    持续时间:215 毫秒和 259 微秒
    与参考解决方案的比较:完全匹配
解决方案“nyarlathotep”:
    持续时间:1341 毫秒和 46 微秒
    与参考解决方案比较:发现不匹配

==================================================== ========

元素数量:1e+06
整数类型的大小:4

参考解决方案:轨道中的轻盈竞赛

解决方案“轨道中的亮度竞赛”:
    持续时间:52861 毫秒和 314 微秒
    与参考解决方案的比较:完全匹配
解决方案“Eric Postpischil”:
    持续时间:4757 毫秒和 608 微秒
    与参考解决方案的比较:完全匹配
解决方案“nyarlathotep”:
    持续时间:15654 毫秒和 195 微秒
    与参考解决方案比较:发现不匹配
解决方案“dyp”:
    持续时间:233 毫秒和 779 微秒
    与参考解决方案的比较:完全匹配
解决方案“pts”:
    持续时间:2181 毫秒和 634 微秒
    与参考解决方案的比较:完全匹配
解决方案“尼姆”:
    持续时间:2539 毫秒和 9 微秒
    与参考解决方案的比较:完全匹配
解决方案“不错”:
    持续时间:2675 毫秒和 362 微秒
    与参考解决方案的比较:完全匹配

==================================================== ========

元素数量:1e+07
整数类型的大小:4

参考解决方案:轨道中的轻盈竞赛

解决方案“不错”:
    持续时间:33425 毫秒和 423 微秒
    与参考解决方案的比较:完全匹配
解决方案“pts”:
    持续时间:26000 毫秒和 398 微秒
    与参考解决方案的比较:完全匹配
解决方案“Eric Postpischil”:
    持续时间:56206 毫秒和 359 微秒
    与参考解决方案的比较:完全匹配
解决方案“轨道中的亮度竞赛”:
    持续时间:658540 毫秒和 342 微秒
    与参考解决方案的比较:完全匹配
解决方案“nyarlathotep”:
    持续时间:187064 毫秒和 518 微秒
    与参考解决方案比较:发现不匹配
解决方案“尼姆”:
    持续时间:30519 毫秒和 227 微秒
    与参考解决方案的比较:完全匹配
解决方案“dyp”:
    持续时间:2624 毫秒和 644 微秒
    与参考解决方案的比较:完全匹配

算法必须是具有支持接口的函数调用运算符模板的结构:

template<class RaIt> operator()(RaIt begin, RaIt end);

输入数据的副本作为参数提供,算法预期提供相同范围内的结果(例如就地排序)。

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <random>
#include <vector>
#include <utility>
#include <cmath>
#include <cassert>
#include <chrono>
#include <cstring>
#include <climits>
#include <functional>
#include <cstdlib>
#include <iomanip>

using duration_t = decltype(  std::chrono::high_resolution_clock::now()
                            - std::chrono::high_resolution_clock::now());

template<class T>
struct result_t
{
    std::vector<T> numbers;
    duration_t duration;
    char const* name;
};

template<class RaIt, class F>
result_t<typename std::iterator_traits<RaIt>::value_type>
apply_algorithm(RaIt p_beg, RaIt p_end, F f, char const* name)
{
    using value_type = typename std::iterator_traits<RaIt>::value_type;

    std::vector<value_type> inplace(p_beg, p_end);

    auto start = std::chrono::high_resolution_clock::now();

    f(begin(inplace), end(inplace));

    auto end = std::chrono::high_resolution_clock::now();
    auto duration = end - start;

    return {std::move(inplace), duration, name};
}

// non-optimized version
int count_digits(int p) // returns `0` for `p == 0`
{
        int res = 0;
        for(; p != 0; ++res)
        {
            p /= 10;
        }
        return res;
}

// non-optimized version
int my_pow10(unsigned exp)
{
        int res = 1;
        for(; exp != 0; --exp)
        {
            res *= 10;
        }
        return res;
}


// !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
// paste algorithms here
// !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


int main(int argc, char** argv)
{
    using integer_t = int;
    constexpr integer_t dist_min = 0;
    constexpr integer_t dist_max = std::numeric_limits<integer_t>::max()/10;
    constexpr std::size_t default_number_of_elements = 1E6;

    const std::size_t number_of_elements = argc>1 ? std::atoll(argv[1]) :
                                           default_number_of_elements;
    std::cout << "number of elements: ";
    std::cout << std::scientific << std::setprecision(0);
    std::cout << (double)number_of_elements << "\n";
    std::cout << /*std::defaultfloat <<*/ std::setprecision(6);
    std::cout.unsetf(std::ios_base::floatfield);

    std::cout << "size of integer type: " << sizeof(integer_t) << "\n\n";

    std::vector<integer_t> input;
    {
        input.reserve(number_of_elements);

        std::random_device rd;
        std::mt19937 gen( rd() );
        std::uniform_int_distribution<> dist(dist_min, dist_max);

        for(std::size_t i = 0; i < number_of_elements; ++i)
            input.push_back( dist(gen) );
    }

    auto b = begin(input);
    auto e = end(input);

    using res_t = result_t<integer_t>;
    std::vector< std::function<res_t()> > algorithms;

    #define MAKE_BINDER(B, E, ALGO, NAME) \
        std::bind( &apply_algorithm<decltype(B),decltype(ALGO)>, \
                   B,E,ALGO,NAME )
    constexpr auto lightness_name = "Lightness Races in Orbit";
    algorithms.push_back( MAKE_BINDER(b, e, lightness(), lightness_name) );
    algorithms.push_back( MAKE_BINDER(b, e, dyp(), "dyp") );
    algorithms.push_back( MAKE_BINDER(b, e, nim(), "Nim") );
    algorithms.push_back( MAKE_BINDER(b, e, pts(), "pts") );
    algorithms.push_back( MAKE_BINDER(b, e, epost(), "Eric Postpischil") );
    algorithms.push_back( MAKE_BINDER(b, e, nyar(), "nyarlathotep") );
    algorithms.push_back( MAKE_BINDER(b, e, notbad(), "notbad") );

    {
        std::srand( std::random_device()() );
        std::random_shuffle(begin(algorithms), end(algorithms));
    }

    std::vector< result_t<integer_t> > res;
    for(auto& algo : algorithms)
        res.push_back( algo() );

    auto reference_solution
        = *std::find_if(begin(res), end(res),
                        [](result_t<integer_t> const& p)
                        { return 0 == std::strcmp(lightness_name, p.name); });
    std::cout << "reference solution: "<<reference_solution.name<<"\n\n";

    for(auto const& e : res)
    {
        std::cout << "solution \""<<e.name<<"\":\n";
        auto ms =
            std::chrono::duration_cast<std::chrono::microseconds>(e.duration);
        std::cout << "\tduration: "<<ms.count()/1000<<" ms and "
                                   <<ms.count()%1000<<" microseconds\n";

        std::cout << "\tcomparison to reference solution: ";
        if(e.numbers.size() != reference_solution.numbers.size())
        {
            std::cout << "ouput count mismatch\n";
            break;
        }

        auto mismatch = std::mismatch(begin(e.numbers), end(e.numbers),
                                      begin(reference_solution.numbers)).first;
        if(end(e.numbers) == mismatch)
        {
            std::cout << "exact match\n";
        }else
        {
            std::cout << "mismatch found\n";
        }
    }
}

当前算法;注意我用全局函数替换了数字计数器和 pow-of-10,所以如果有人优化,我们都会受益。

struct lightness
{
    template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        using T = typename std::iterator_traits<RaIt>::value_type;

        /**
         * Sorts the array lexicographically.
         *
         * The trick is that we have to compare digits left-to-right
         * (considering typical Latin decimal notation) and that each of
         * two numbers to compare may have a different number of digits.
         *
         * This is very efficient in storage space, but inefficient in
         * execution time; an approach that pre-visits each element and
         * stores a translated representation will at least double your
         * storage requirements (possibly a problem with large inputs)
         * but require only a single translation of each element.
         */
        std::sort(
            b,
            e,
            [](T lhs, T rhs) -> bool {
                // Returns true if lhs < rhs
                // Returns false otherwise
                const auto BASE      = 10;
                const bool LHS_FIRST = true;
                const bool RHS_FIRST = false;
                const bool EQUAL     = false;


                // There's no point in doing anything at all
                // if both inputs are the same; strict-weak
                // ordering requires that we return `false`
                // in this case.
                if (lhs == rhs) {
                    return EQUAL;
                }

                // Compensate for sign
                if (lhs < 0 && rhs < 0) {
                    // When both are negative, sign on its own yields
                    // no clear ordering between the two arguments.
                    //
                    // Remove the sign and continue as for positive
                    // numbers.
                    lhs *= -1;
                    rhs *= -1;
                }
                else if (lhs < 0) {
                    // When the LHS is negative but the RHS is not,
                    // consider the LHS "first" always as we wish to
                    // prioritise the leading '-'.
                    return LHS_FIRST;
                }
                else if (rhs < 0) {
                    // When the RHS is negative but the LHS is not,
                    // consider the RHS "first" always as we wish to
                    // prioritise the leading '-'.
                    return RHS_FIRST;
                }

                // Counting the number of digits in both the LHS and RHS
                // arguments is *almost* trivial.
                const auto lhs_digits = (
                    lhs == 0
                    ? 1
                    : std::ceil(std::log(lhs+1)/std::log(BASE))
                );

                const auto rhs_digits = (
                    rhs == 0
                    ? 1
                    : std::ceil(std::log(rhs+1)/std::log(BASE))
                );

                // Now we loop through the positions, left-to-right,
                // calculating the digit at these positions for each
                // input, and comparing them numerically. The
                // lexicographic nature of the sorting comes from the
                // fact that we are doing this per-digit comparison
                // rather than considering the input value as a whole.
                const auto max_pos = std::max(lhs_digits, rhs_digits);
                for (auto pos = 0; pos < max_pos; pos++) {
                    if (lhs_digits - pos == 0) {
                        // Ran out of digits on the LHS;
                        // prioritise the shorter input
                        return LHS_FIRST;
                    }
                    else if (rhs_digits - pos == 0) {
                        // Ran out of digits on the RHS;
                        // prioritise the shorter input
                        return RHS_FIRST;
                    }
                    else {
                        const auto lhs_x = (lhs / static_cast<decltype(BASE)>(std::pow(BASE, lhs_digits - 1 - pos))) % BASE;
                        const auto rhs_x = (rhs / static_cast<decltype(BASE)>(std::pow(BASE, rhs_digits - 1 - pos))) % BASE;

                        if (lhs_x < rhs_x)
                            return LHS_FIRST;
                        else if (rhs_x < lhs_x)
                            return RHS_FIRST;
                    }
                }

                // If we reached the end and everything still
                // matches up, then something probably went wrong
                // as I'd have expected to catch this in the tests
                // for equality.
                assert("Unknown case encountered");

                // dyp: suppress warning and throw
                throw "up";
            }
        );
    }
};

namespace ndyp
{
    // helper to provide integers with the same number of digits
    template<class T, class U>
    std::pair<T, T> lexicographic_pair_helper(T const p, U const maxDigits)
    {
        auto const digits = count_digits(p);
        // append zeros so that `l` has `maxDigits` digits
        auto const l = static_cast<T>( p  * my_pow10(maxDigits-digits) );
        return {l, p};
    }

    template<class RaIt>
    using pair_vec
        = std::vector<std::pair<typename std::iterator_traits<RaIt>::value_type,
                                typename std::iterator_traits<RaIt>::value_type>>;

    template<class RaIt>
    pair_vec<RaIt> lexicographic_sort(RaIt p_beg, RaIt p_end)
    {
        if(p_beg == p_end) return pair_vec<RaIt>{};

        auto max = *std::max_element(p_beg, p_end);
        auto maxDigits = count_digits(max);

        pair_vec<RaIt> result;
        result.reserve( std::distance(p_beg, p_end) );

        for(auto i = p_beg; i != p_end; ++i)
            result.push_back( lexicographic_pair_helper(*i, maxDigits) );

        using value_type = typename pair_vec<RaIt>::value_type;

        std::sort(begin(result), end(result),
                  [](value_type const& l, value_type const& r)
                  {
                      if(l.first < r.first) return true;
                      if(l.first > r.first) return false;
                      return l.second < r.second; }
                 );

        return result;
    }
}

struct dyp
{
    template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        auto pairvec = ndyp::lexicographic_sort(b, e);
        std::transform(begin(pairvec), end(pairvec), b,
                       [](typename decltype(pairvec)::value_type const& e) { return e.second; });
    }
};


namespace nnim
{
    bool comp(int l, int r)
    {
      int lv[10] = {}; // probably possible to get this from numeric_limits
      int rv[10] = {};

      int lc = 10; // ditto
      int rc = 10;
      while (l || r)
      {
        if (l)
        {
          auto t = l / 10;
          lv[--lc] = l - (t * 10);
          l = t;
        }
        if (r)
        {
          auto t = r / 10;
          rv[--rc] = r - (t * 10);
          r = t;
        }
      }
      while (lc < 10 && rc < 10)
      {
        if (lv[lc] == rv[rc])
        {
          lc++;
          rc++;
        }
        else
          return lv[lc] < rv[rc];
      }
      return lc > rc;
    }
}

struct nim
{
    template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, nnim::comp);
    }
};

struct pts
{
        template<class T> static bool lex_less(T a, T b) {
          unsigned la = 1, lb = 1;
          for (T t = a; t > 9; t /= 10) ++la;
          for (T t = b; t > 9; t /= 10) ++lb;
          const bool ll = la < lb;
          while (la > lb) { b *= 10; ++lb; }
          while (lb > la) { a *= 10; ++la; }
          return a == b ? ll : a < b;
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, lex_less<typename std::iterator_traits<RaIt>::value_type>);
    }
};

struct epost
{
        static bool compare(int x, int y)
        {
                static const double limit = .5 * (log(INT_MAX) - log(INT_MAX-1));

                double lx = log10(x);
                double ly = log10(y);
                double fx = lx - floor(lx);  // Get the mantissa of lx.
                double fy = ly - floor(ly);  // Get the mantissa of ly.
                return fabs(fx - fy) < limit ? lx < ly : fx < fy;
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, compare);
    }
};

struct nyar
{
        static bool lexiSmaller(int i1, int i2)
        {
            int digits1 = count_digits(i1);
            int digits2 = count_digits(i2);

            double val1 = i1/pow(10.0, digits1-1);
            double val2 = i2/pow(10.0, digits2-1);

            while (digits1 > 0 && digits2 > 0 && (int)val1 == (int)val2)
            {
                digits1--;
                digits2--;
                val1 = (val1 - (int)val1)*10;
                val2 = (val2 - (int)val2)*10;
            }
            if (digits1 > 0 && digits2 > 0)
            {
                return (int)val1 < (int)val2;
            }
            return (digits2 > 0);
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, lexiSmaller);
    }
};

struct notbad
{
        static int up_10pow(int n) {
          int ans = 1;
          while (ans < n) ans *= 10;
          return ans;
        }

        static bool compare(int v1, int v2) {
           int ceil1 = up_10pow(v1), ceil2 = up_10pow(v2);
           while ( ceil1 != 0 && ceil2 != 0) {
              if (v1 / ceil1  < v2 / ceil2) return true;
              else if (v1 / ceil1 > v2 / ceil2) return false;
              ceil1 /= 10;
              ceil2 /= 10;
           }
           if (v1 < v2) return true;
           return false;
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, compare);
    }
};
于 2013-10-25T18:47:42.903 回答
6

这是另一种算法,它在排序之前进行一些计算。尽管有额外的复制,但它似乎相当快(参见比较)。

笔记:

  • 它只支持正整数
  • in 仅支持整数 <=std::numeric_limits<int>::max()/10

注意你可以优化count_digitsmy_pow10;例如,请参阅Andrei Alexandrescu 的 C++ 的三个优化技巧任何比 pow() 更快地在 C++ 中计算 10 的整数幂的方法?

帮手:

#include <random>
#include <vector>
#include <utility>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <limits>
#include <iterator>

// non-optimized version
int count_digits(int p) // returns `0` for `p == 0`
{
    int res = 0;
    for(; p != 0; ++res)
    {
        p /= 10;
    }
    return res;
}

// non-optimized version
int my_pow10(unsigned exp)
{
    int res = 1;
    for(; exp != 0; --exp)
    {
        res *= 10;
    }
    return res;
}

算法(注意 -不是就地):

// helper to provide integers with the same number of digits
template<class T, class U>
std::pair<T, T> lexicographic_pair_helper(T const p, U const maxDigits)
{
    auto const digits = count_digits(p);
    // append zeros so that `l` has `maxDigits` digits
    auto const l = static_cast<T>( p  * my_pow10(maxDigits-digits) );
    return {l, p};
}

template<class RaIt>
using pair_vec
    = std::vector<std::pair<typename std::iterator_traits<RaIt>::value_type,
                            typename std::iterator_traits<RaIt>::value_type>>;

template<class RaIt>
pair_vec<RaIt> lexicographic_sort(RaIt p_beg, RaIt p_end)
{
    if(p_beg == p_end) return {};

    auto max = *std::max_element(p_beg, p_end);
    auto maxDigits = count_digits(max);

    pair_vec<RaIt> result;
    result.reserve( std::distance(p_beg, p_end) );

    for(auto i = p_beg; i != p_end; ++i)
        result.push_back( lexicographic_pair_helper(*i, maxDigits) );

    using value_type = typename pair_vec<RaIt>::value_type;

    std::sort(begin(result), end(result),
              [](value_type const& l, value_type const& r)
              {
                  if(l.first < r.first) return true;
                  if(l.first > r.first) return false;
                  return l.second < r.second; }
             );

    return result;
}

使用示例:

int main()
{
    std::vector<int> input = { 100, 21 , 22 , 99 , 1  , 927 };
    // generate some numbers
    /*{
        constexpr int number_of_elements = 1E6;
        std::random_device rd;
        std::mt19937 gen( rd() );
        std::uniform_int_distribution<>
            dist(0, std::numeric_limits<int>::max()/10);
        for(int i = 0; i < number_of_elements; ++i)
            input.push_back( dist(gen) );
    }*/

    std::cout << "unsorted: ";
    for(auto const& e : input) std::cout << e << ", ";
    std::cout << "\n\n";


    auto sorted = lexicographic_sort(begin(input), end(input));

    std::cout << "sorted: ";
    for(auto const& e : sorted) std::cout << e.second << ", ";
    std::cout << "\n\n";
}
于 2013-10-25T13:01:26.997 回答
4

我相信以下作为正整数的排序比较函数,只要使用的整数类型比double类型(例如, 32-bitint和 64-bit double)窄得多,并且log10使用的例程返回精确的 10 次幂的完全正确的结果(其中一个好的实现确实):

static const double limit = .5 * (log(INT_MAX) - log(INT_MAX-1));

double lx = log10(x);
double ly = log10(y);
double fx = lx - floor(lx);  // Get the mantissa of lx.
double fy = ly - floor(ly);  // Get the mantissa of ly.
return fabs(fx - fy) < limit ? lx < ly : fx < fy;

它通过比较对数的尾数来工作。尾数是对数的小数部分,它们表示没有大小的数字的有效数字的值(例如,31、3.1 和 310 的对数具有完全相同的尾数)。

的目的fabs(fx - fy) < limit是允许在取对数时出现错误,这既是因为 的实现log10不完善,也因为浮点格式会导致一些错误。(31 和 310 的对数的整数部分使用不同的位数,因此有效位数的位数不同,因此它们最终会被舍入到稍微不同的值。)只要整数类型明显更窄比double类型,计算出来limit的误差会远大于log10. 因此,该测试fabs(fx - fy) < limit基本上告诉我们如果精确计算,两个计算的尾数是否相等。

如果尾数不同,它们表示字典顺序,所以我们返回fx < fy。如果它们相等,那么对数的整数部分告诉我们顺序,所以我们返回lx < ly

很容易测试是否log10为每个 10 的幂返回正确的结果,因为它们太少了。如果没有,可以轻松进行调整:插入if (1-fx < limit) fx = 0; if (1-fu < limit) fy = 0;。这允许log10当它应该返回 5 时返回类似 4.99999 的东西。

这种方法的优点是不使用循环或除法(这在许多处理器上很耗时)。

于 2013-10-25T13:44:19.427 回答
3

如果您的所有数字都是非负数并且它们足够小以至于将它们乘以 10 不会导致溢出,则这是一个紧凑的解决方案:

template<class T> bool lex_less(T a, T b) {
  unsigned la = 1, lb = 1;
  for (T t = a; t > 9; t /= 10) ++la;
  for (T t = b; t > 9; t /= 10) ++lb;
  const bool ll = la < lb;
  while (la > lb) { b *= 10; ++lb; }
  while (lb > la) { a *= 10; ++la; }
  return a == b ? ll : a < b;
}

像这样运行它:

#include <iostream>
#include <algorithm>
int main(int, char **) {
  unsigned short input[] = { 100, 21 , 22 , 99 , 1  , 927 };
  unsigned input_size = sizeof(input) / sizeof(input[0]);
  std::sort(input, input + input_size, lex_less<unsigned short>);
  for (unsigned i = 0; i < input_size; ++i) {
    std::cout << ' ' << input[i];
  }
  std::cout << std::endl;
  return 0;
}
于 2013-10-25T16:57:52.523 回答
3

这项任务听起来很适合带有填充的 Radix Sort 的 MSD 变体 ( http://en.wikipedia.org/wiki/Radix_sort )。

取决于你想扔多少代码。其他人显示的简单代码是 O(log n) 复杂度,而完全优化的基数排序是 O(kn)。

于 2013-10-25T13:14:01.200 回答
1

您可以尝试使用 % 运算符让您访问每个单独的数字,例如 121 % 100 会给您第一个数字并以这种方式检查,但您必须找到一种方法来解决它们具有不同大小的事实。

所以找到数组中的最大值。我不知道您是否可以尝试内置此功能。

int Max (int* pdata,int size)
 {
int temp_max =0 ;

for (int i =0 ; i < size ; i++)
 {
    if (*(pdata+i) > temp_max)
    {
        temp_max = *(pdata+i);

    }
 }
 return temp_max;
 }

此函数将返回数字中的位数

 int Digit_checker(int n)
{
 int num_digits = 1;

while (true)
{
    if ((n % 10) == n)
        return num_digits;
    num_digits++;
    n = n/10;
}
return num_digits;
}

令 max 中的位数等于 n。一旦你以 for (int i = 1; i < n ; i++) 的格式打开一个 for 循环

然后你可以通过你的并使用“data[i] % (10^(ni))”来访问第一个数字然后对其进行排序,然后在下一次迭代中你将可以访问第二个数字。我不知道你会如何排序它们。

它不适用于负数,您必须绕过 data[i] % (10^(ni)) 为位数少于最大值的数字返回自身

于 2013-10-25T11:51:24.070 回答
1

重载 < 运算符以按字典顺序比较两个整数。对于每个整数,找到不小于给定整数的最小 10^k。而不是一一比较数字。

class CmpIntLex {
int up_10pow(int n) {
  int ans = 1;
  while (ans < n) ans *= 10;
  return ans;
}
public: 
bool operator ()(int v1, int v2) {
   int ceil1 = up_10pow(v1), ceil2 = up_10pow(v2);
   while ( ceil1 != 0 && ceil2 != 0) {
      if (v1 / ceil1  < v2 / ceil2) return true;
      else if (v1 / ceil1 > v2 / ceil2) return false;
      ceil1 /= 10; 
      ceil2 /= 10;
   }
   if (v1 < v2) return true;
   return false;
}
int main() {
vector<int> vi = {12,45,12134,85};
sort(vi.begin(), vi.end(), CmpIntLex());
}
于 2013-10-25T12:06:59.990 回答
0

虽然这里的其他一些答案(Lightness's, notbad's)已经显示出相当不错的代码,但我相信我可以添加一个性能更高的解决方案(因为它在每个循环中既不需要除法也不需要幂;但它需要浮点运算,这又是可能会使它变慢,并且对于大量数据可能不准确):

#include <algorithm>
#include <iostream>
#include <assert.h>

// method taken from http://stackoverflow.com/a/1489873/671366
template <class T>
int numDigits(T number)
{
    int digits = 0;
    if (number < 0) digits = 1; // remove this line if '-' counts as a digit
    while (number) {
        number /= 10;
        digits++;
    }
    return digits;
}

bool lexiSmaller(int i1, int i2)
{
    int digits1 = numDigits(i1);
    int digits2 = numDigits(i2);

    double val1 = i1/pow(10.0, digits1-1);
    double val2 = i2/pow(10.0, digits2-1);

    while (digits1 > 0 && digits2 > 0 && (int)val1 == (int)val2)
    {
        digits1--;
        digits2--;
        val1 = (val1 - (int)val1)*10;
        val2 = (val2 - (int)val2)*10;
    }
    if (digits1 > 0 && digits2 > 0)
    {
        return (int)val1 < (int)val2;
    }
    return (digits2 > 0);
}


int main(int argc, char* argv[])
{
    // just testing whether the comparison function works as expected:
    assert (lexiSmaller(1, 100));
    assert (!lexiSmaller(100, 1));
    assert (lexiSmaller(100, 22));
    assert (!lexiSmaller(22, 100));
    assert (lexiSmaller(927, 99));
    assert (!lexiSmaller(99, 927));
    assert (lexiSmaller(1, 927));
    assert (!lexiSmaller(927, 1));
    assert (lexiSmaller(21, 22));
    assert (!lexiSmaller(22, 21));
    assert (lexiSmaller(22, 99));
    assert (!lexiSmaller(99, 22));

    // use the comparison function for the actual sorting:
    int input[] = { 100 , 21 , 22 , 99 , 1 ,927 };
    std::sort(&input[0], &input[5], lexiSmaller);
    std::cout << "sorted: ";
    for (int i=0; i<6; ++i)
    {
        std::cout << input[i];
        if (i<5)
        {
            std::cout << ", ";
        }
    }
    std::cout << std::endl;
    return 0;
}

虽然我不得不承认我还没有测试过性能。

于 2013-10-25T12:13:09.700 回答
0

这是不使用任何浮点技巧的愚蠢解决方案。它与字符串比较几乎相同,但不使用字符串,也不处理负数,为此在顶部添加一个部分......

bool comp(int l, int r)
{
  int lv[10] = {}; // probably possible to get this from numeric_limits
  int rv[10] = {};

  int lc = 10; // ditto
  int rc = 10;
  while (l || r)
  {
    if (l)
    {
      auto t = l / 10;
      lv[--lc] = l - (t * 10);
      l = t;
    }
    if (r)
    {
      auto t = r / 10;
      rv[--rc] = r - (t * 10);
      r = t;
    }
  }
  while (lc < 10 && rc < 10)
  {
    if (lv[lc] == rv[rc])
    {
      lc++;
      rc++;
    }
    else
      return lv[lc] < rv[rc];
  }
  return lc > rc;
}

它很快,而且我敢肯定还有可能让它更快,但它确实有效,而且它很愚蠢,无法理解......

编辑:我吃了很多代码,但这是迄今为止所有解决方案的比较..

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <random>
#include <vector>
#include <utility>
#include <cmath>
#include <cassert>
#include <chrono>

std::pair<int, int> lexicographic_pair_helper(int p, int maxDigits)
{
  int digits = std::log10(p);
  int l = p*std::pow(10, maxDigits-digits);
  return {l, p};
}

bool l_comp(int l, int r)
{
  int lv[10] = {}; // probably possible to get this from numeric_limits
  int rv[10] = {};

  int lc = 10; // ditto
  int rc = 10;
  while (l || r)
  {
    if (l)
    {
      auto t = l / 10;
      lv[--lc] = l - (t * 10);
      l = t;
    }
    if (r)
    {
      auto t = r / 10;
      rv[--rc] = r - (t * 10);
      r = t;
    }
  }
  while (lc < 10 && rc < 10)
  {
    if (lv[lc] == rv[rc])
    {
      lc++;
      rc++;
    }
    else
      return lv[lc] < rv[rc];
  }
  return lc > rc;
}

int up_10pow(int n) {
  int ans = 1;
  while (ans < n) ans *= 10;
  return ans;
}
bool l_comp2(int v1, int v2) {
  int n1 = up_10pow(v1), n2 = up_10pow(v2);
  while ( v1 != 0 && v2 != 0) {
    if (v1 / n1  < v2 / n2) return true;
    else if (v1 / n1 > v2 / n2) return false;
    v1 /= 10;
    v2 /= 10;
    n1 /= 10;
    n2 /= 10;
  }
  if (v1 == 0 && v2 != 0) return true;
  return false;
}

int main()
{
  std::vector<int> numbers;
  {
    constexpr int number_of_elements = 1E6;
    std::random_device rd;
    std::mt19937 gen( rd() );
    std::uniform_int_distribution<> dist;
    for(int i = 0; i < number_of_elements; ++i) numbers.push_back( dist(gen) );
  }

  std::vector<int> lo(numbers);
  std::vector<int> dyp(numbers);
  std::vector<int> nim(numbers);
  std::vector<int> nb(numbers);

  std::cout << "starting..." << std::endl;

  {

    auto start = std::chrono::high_resolution_clock::now();
    /**
    * Sorts the array lexicographically.
    *
    * The trick is that we have to compare digits left-to-right
    * (considering typical Latin decimal notation) and that each of
    * two numbers to compare may have a different number of digits.
    *
    * This probably isn't very efficient, so I wouldn't do it on
    * "millions" of numbers. But, it works...
    */
    std::sort(
    std::begin(lo),
              std::end(lo),
              [](int lhs, int rhs) -> bool {
                // Returns true if lhs < rhs
                // Returns false otherwise
                const auto BASE      = 10;
                const bool LHS_FIRST = true;
                const bool RHS_FIRST = false;
                const bool EQUAL     = false;


                // There's no point in doing anything at all
                // if both inputs are the same; strict-weak
                // ordering requires that we return `false`
                // in this case.
                if (lhs == rhs) {
                  return EQUAL;
                }

                // Compensate for sign
                if (lhs < 0 && rhs < 0) {
                  // When both are negative, sign on its own yields
                  // no clear ordering between the two arguments.
                  //
                  // Remove the sign and continue as for positive
                  // numbers.
                  lhs *= -1;
                  rhs *= -1;
                }
                else if (lhs < 0) {
                  // When the LHS is negative but the RHS is not,
              // consider the LHS "first" always as we wish to
              // prioritise the leading '-'.
              return LHS_FIRST;
                }
                else if (rhs < 0) {
                  // When the RHS is negative but the LHS is not,
              // consider the RHS "first" always as we wish to
              // prioritise the leading '-'.
              return RHS_FIRST;
                }

                // Counting the number of digits in both the LHS and RHS
                // arguments is *almost* trivial.
                const auto lhs_digits = (
                lhs == 0
                ? 1
                : std::ceil(std::log(lhs+1)/std::log(BASE))
                );

                const auto rhs_digits = (
                rhs == 0
                ? 1
                : std::ceil(std::log(rhs+1)/std::log(BASE))
                );

                // Now we loop through the positions, left-to-right,
              // calculating the digit at these positions for each
              // input, and comparing them numerically. The
              // lexicographic nature of the sorting comes from the
              // fact that we are doing this per-digit comparison
              // rather than considering the input value as a whole.
              const auto max_pos = std::max(lhs_digits, rhs_digits);
              for (auto pos = 0; pos < max_pos; pos++) {
                if (lhs_digits - pos == 0) {
                  // Ran out of digits on the LHS;
                  // prioritise the shorter input
                  return LHS_FIRST;
                }
                else if (rhs_digits - pos == 0) {
                  // Ran out of digits on the RHS;
                  // prioritise the shorter input
                  return RHS_FIRST;
                }
                else {
                  const auto lhs_x = (lhs / static_cast<decltype(BASE)>(std::pow(BASE, lhs_digits - 1 - pos))) % BASE;
                  const auto rhs_x = (rhs / static_cast<decltype(BASE)>(std::pow(BASE, rhs_digits - 1 - pos))) % BASE;

                  if (lhs_x < rhs_x)
                    return LHS_FIRST;
                  else if (rhs_x < lhs_x)
                    return RHS_FIRST;
                }
              }

              // If we reached the end and everything still
              // matches up, then something probably went wrong
              // as I'd have expected to catch this in the tests
              // for equality.
              assert("Unknown case encountered");
              }
              );

    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = end - start;
    std::cout << "Lightness: " << elapsed.count() << '\n';
  }

  {
    auto start = std::chrono::high_resolution_clock::now();

    auto max = *std::max_element(begin(dyp), end(dyp));
    int maxDigits = std::log10(max);

    std::vector<std::pair<int,int>> temp;
    temp.reserve(dyp.size());
    for(auto const& e : dyp) temp.push_back( lexicographic_pair_helper(e, maxDigits) );

    std::sort(begin(temp), end(temp), [](std::pair<int, int> const& l, std::pair<int, int> const& r)
    { if(l.first < r.first) return true; if(l.first > r.first) return false; return l.second < r.second; });

    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = end - start;
    std::cout << "Dyp: " << elapsed.count() << '\n';
  }

  {
    auto start = std::chrono::high_resolution_clock::now();
    std::sort (nim.begin(), nim.end(), l_comp);
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = end - start;
    std::cout << "Nim: " << elapsed.count() << '\n';
  }

//   {
//     auto start = std::chrono::high_resolution_clock::now();
//     std::sort (nb.begin(), nb.end(), l_comp2);
//     auto end = std::chrono::high_resolution_clock::now();
//     auto elapsed = end - start;
//     std::cout << "notbad: " << elapsed.count() << '\n';
//   }

  std::cout << (nim == lo) << std::endl;
  std::cout << (nim == dyp) << std::endl;
  std::cout << (lo == dyp) << std::endl;
//   std::cout << (lo == nb) << std::endl;
}
于 2013-10-25T15:55:47.970 回答
0

根据@Oswald 的回答,下面是一些执行相同操作的代码。

#include <iostream>
#include <vector>
#include <algorithm> 
using namespace std;

bool compare(string a, string b){
    // Check each digit
    int i = 0, j = 0;
    while(i < a.size() && j < b.size()){
        // If different digits
        if(a[i] - '0' != b[j] - '0')
            return (a[i] - '0' < b[j] - '0');
        i++, j++;
    }
    // Different sizes
    return (a.size() < b.size());
}

int main(){
    vector<string> array = {"1","2","3","4","5","6","7","8","9","10","11","12"};
    sort(array.begin(), array.end(), compare);

    for(auto value : array)
        cout << value << " ";
    return 0;
}

输入:1 2 3 4 5 6 7 8 9 10 11 12

输出:1 10 11 12 2 3 4 5 6 7 8 9

于 2018-11-05T05:17:52.013 回答