我需要获取一个可能包含很多元素的 C++ 向量，删除重复项并对其进行排序。
我目前有以下代码，但它不起作用。
vec.erase(
std::unique(vec.begin(), vec.end()),
vec.end());
std::sort(vec.begin(), vec.end());
我怎样才能正确地做到这一点？
此外，先删除重复项（类似于上面的代码）还是先执行排序更快？如果我确实先执行排序，是否保证在std::unique执行后保持排序？
还是有另一种（可能更有效）的方式来完成这一切？
我同意R. Pate和Todd Gardner的观点；astd::set在这里可能是个好主意。即使您无法使用向量，但如果您有足够的重复项，您最好创建一个集合来完成这些繁琐的工作。
让我们比较三种方法：
仅使用向量，排序+唯一
sort( vec.begin(), vec.end() );
vec.erase( unique( vec.begin(), vec.end() ), vec.end() );
转换为设置（手动）
set<int> s;
unsigned size = vec.size();
for( unsigned i = 0; i < size; ++i ) s.insert( vec[i] );
vec.assign( s.begin(), s.end() );
转换为集合（使用构造函数）
set<int> s( vec.begin(), vec.end() );
vec.assign( s.begin(), s.end() );
以下是这些随着重复数量的变化而变化的表现：
摘要：当重复的数量足够大时，转换为集合然后将数据转储回向量实际上更快。
出于某种原因，手动进行集合转换似乎比使用集合构造函数更快——至少在我使用的玩具随机数据上是这样。
我重做了 Nate Kohl 的分析并得到了不同的结果。对于我的测试用例，直接对向量进行排序总是比使用集合更有效。我添加了一种新的更有效的方法，使用unordered_set.
请记住，unordered_set只有当您对需要唯一和排序的类型具有良好的散列函数时，该方法才有效。对于整数，这很容易！（标准库提供了一个默认的哈希，它只是身份函数。）另外，不要忘记在最后进行排序，因为 unordered_set 是无序的:)
我在setandunordered_set实现中进行了一些挖掘，发现构造函数实际上为每个元素构造了一个新节点，然后检查它的值以确定它是否应该实际插入（至少在 Visual Studio 实现中）。
以下是5种方法：
f1: 只需使用vector, sort+unique
sort( vec.begin(), vec.end() );
vec.erase( unique( vec.begin(), vec.end() ), vec.end() );
f2：转换为set（使用构造函数）
set<int> s( vec.begin(), vec.end() );
vec.assign( s.begin(), s.end() );
f3：转换为set（手动）
set<int> s;
for (int i : vec)
s.insert(i);
vec.assign( s.begin(), s.end() );
f4：转换为unordered_set（使用构造函数）
unordered_set<int> s( vec.begin(), vec.end() );
vec.assign( s.begin(), s.end() );
sort( vec.begin(), vec.end() );
f5：转换为unordered_set（手动）
unordered_set<int> s;
for (int i : vec)
s.insert(i);
vec.assign( s.begin(), s.end() );
sort( vec.begin(), vec.end() );
我使用在 [1,10]、[1,1000] 和 [1,100000] 范围内随机选择的 100,000,000 个整数的向量进行了测试
结果（以秒为单位，越小越好）：
range f1 f2 f3 f4 f5
[1,10] 1.6821 7.6804 2.8232 6.2634 0.7980
[1,1000] 5.0773 13.3658 8.2235 7.6884 1.9861
[1,100000] 8.7955 32.1148 26.5485 13.3278 3.9822
std::unique如果它们是邻居，则仅删除重复元素：您必须先对向量进行排序，然后它才能按预期工作。
std::unique被定义为稳定的，所以向量在运行 unique 后仍然会被排序。
我不确定你在用这个做什么，所以我不能 100% 肯定地说这个，但通常当我想到“排序的、唯一的”容器时，我会想到一个std::set。它可能更适合您的用例：
std::set<Foo> foos(vec.begin(), vec.end()); // both sorted & unique already
否则，在调用唯一之前进行排序（正如其他答案所指出的那样）是要走的路。
std::unique仅适用于重复元素的连续运行，因此您最好先排序。但是，它是稳定的，因此您的向量将保持排序状态。
这是一个为您做的模板：
template<typename T>
void removeDuplicates(std::vector<T>& vec)
{
std::sort(vec.begin(), vec.end());
vec.erase(std::unique(vec.begin(), vec.end()), vec.end());
}
像这样称呼它：
removeDuplicates<int>(vectorname);
如果您不想更改元素的顺序，则可以尝试以下解决方案：
template <class T>
void RemoveDuplicatesInVector(std::vector<T> & vec)
{
set<T> values;
vec.erase(std::remove_if(vec.begin(), vec.end(), [&](const T & value) { return !values.insert(value).second; }), vec.end());
}
效率是一个复杂的概念。有时间与空间的考虑，以及一般测量（您只能得到模糊的答案，例如 O(n)）与特定的测量（例如，冒泡排序可能比快速排序快得多，具体取决于输入特征）。
如果您的重复项相对较少，那么排序后跟唯一并擦除似乎是要走的路。如果您有相对较多的重复项，则从向量创建一个集合并让它完成繁重的工作可以轻松击败它。
也不要只关注时间效率。排序+唯一+擦除在 O(1) 空间中运行，而集合构造在 O(n) 空间中运行。并且两者都不直接适用于 map-reduce 并行化（对于非常大的数据集）。
假设a是一个向量，使用
a.erase(unique(a.begin(),a.end()),a.end());在O(n)时间内运行。
您需要在调用之前对其进行排序，unique因为unique只删除彼此相邻的重复项。
编辑：38秒...
unique只删除连续的重复元素（这是在线性时间内运行所必需的），因此您应该先执行排序。调用后它将保持排序状态unique。
使用 Ranges v3 库，您可以简单地使用
action::unique(vec);
请注意，它实际上删除了重复的元素，而不仅仅是移动它们。
不幸的是，C++20 中的操作并未标准化，因为范围库的其他部分即使在 C++20 中您仍然必须使用原始库。
如前所述，unique需要一个分类容器。此外，unique实际上并不从容器中删除元素。相反，它们被复制到最后，unique返回一个指向第一个这样的重复元素的迭代器，并且您应该调用erase以实际删除元素。
您可以按如下方式执行此操作：
std::sort(v.begin(), v.end());
v.erase(std::unique(v.begin(), v.end()), v.end());
Nate Kohl 建议的标准方法，仅使用向量、排序 + 唯一：
sort( vec.begin(), vec.end() );
vec.erase( unique( vec.begin(), vec.end() ), vec.end() );
不适用于指针向量。
仔细查看cplusplus.com 上的这个示例。
在他们的示例中，移到末尾的“所谓的重复项”实际上显示为？（未定义的值），因为那些“所谓的重复项”有时是“额外元素”，有时在原始向量中存在“缺失元素”。
在指向对象的指针向量上使用时会出现问题std::unique()（内存泄漏、从 HEAP 读取数据错误、重复释放，这会导致分段错误等）。
这是我对问题的解决方案：替换std::unique()为ptgi::unique().
请参阅下面的文件 ptgi_unique.hpp：
// ptgi::unique()
//
// Fix a problem in std::unique(), such that none of the original elts in the collection are lost or duplicate.
// ptgi::unique() has the same interface as std::unique()
//
// There is the 2 argument version which calls the default operator== to compare elements.
//
// There is the 3 argument version, which you can pass a user defined functor for specialized comparison.
//
// ptgi::unique() is an improved version of std::unique() which doesn't looose any of the original data
// in the collection, nor does it create duplicates.
//
// After ptgi::unique(), every old element in the original collection is still present in the re-ordered collection,
// except that duplicates have been moved to a contiguous range [dupPosition, last) at the end.
//
// Thus on output:
// [begin, dupPosition) range are unique elements.
// [dupPosition, last) range are duplicates which can be removed.
// where:
// [] means inclusive, and
// () means exclusive.
//
// In the original std::unique() non-duplicates at end are moved downward toward beginning.
// In the improved ptgi:unique(), non-duplicates at end are swapped with duplicates near beginning.
//
// In addition if you have a collection of ptrs to objects, the regular std::unique() will loose memory,
// and can possibly delete the same pointer multiple times (leading to SEGMENTATION VIOLATION on Linux machines)
// but ptgi::unique() won't. Use valgrind(1) to find such memory leak problems!!!
//
// NOTE: IF you have a vector of pointers, that is, std::vector<Object*>, then upon return from ptgi::unique()
// you would normally do the following to get rid of the duplicate objects in the HEAP:
//
// // delete objects from HEAP
// std::vector<Object*> objects;
// for (iter = dupPosition; iter != objects.end(); ++iter)
// {
// delete (*iter);
// }
//
// // shrink the vector. But Object * pointers are NOT followed for duplicate deletes, this shrinks the vector.size())
// objects.erase(dupPosition, objects.end));
//
// NOTE: But if you have a vector of objects, that is: std::vector<Object>, then upon return from ptgi::unique(), it
// suffices to just call vector:erase(, as erase will automatically call delete on each object in the
// [dupPosition, end) range for you:
//
// std::vector<Object> objects;
// objects.erase(dupPosition, last);
//
//==========================================================================================================
// Example of differences between std::unique() vs ptgi::unique().
//
// Given:
// int data[] = {10, 11, 21};
//
// Given this functor: ArrayOfIntegersEqualByTen:
// A functor which compares two integers a[i] and a[j] in an int a[] array, after division by 10:
//
// // given an int data[] array, remove consecutive duplicates from it.
// // functor used for std::unique (BUGGY) or ptgi::unique(IMPROVED)
//
// // Two numbers equal if, when divided by 10 (integer division), the quotients are the same.
// // Hence 50..59 are equal, 60..69 are equal, etc.
// struct ArrayOfIntegersEqualByTen: public std::equal_to<int>
// {
// bool operator() (const int& arg1, const int& arg2) const
// {
// return ((arg1/10) == (arg2/10));
// }
// };
//
// Now, if we call (problematic) std::unique( data, data+3, ArrayOfIntegersEqualByTen() );
//
// TEST1: BEFORE UNIQ: 10,11,21
// TEST1: AFTER UNIQ: 10,21,21
// DUP_INX=2
//
// PROBLEM: 11 is lost, and extra 21 has been added.
//
// More complicated example:
//
// TEST2: BEFORE UNIQ: 10,20,21,22,30,31,23,24,11
// TEST2: AFTER UNIQ: 10,20,30,23,11,31,23,24,11
// DUP_INX=5
//
// Problem: 21 and 22 are deleted.
// Problem: 11 and 23 are duplicated.
//
//
// NOW if ptgi::unique is called instead of std::unique, both problems go away:
//
// DEBUG: TEST1: NEW_WAY=1
// TEST1: BEFORE UNIQ: 10,11,21
// TEST1: AFTER UNIQ: 10,21,11
// DUP_INX=2
//
// DEBUG: TEST2: NEW_WAY=1
// TEST2: BEFORE UNIQ: 10,20,21,22,30,31,23,24,11
// TEST2: AFTER UNIQ: 10,20,30,23,11,31,22,24,21
// DUP_INX=5
//
// @SEE: look at the "case study" below to understand which the last "AFTER UNIQ" results with that order:
// TEST2: AFTER UNIQ: 10,20,30,23,11,31,22,24,21
//
//==========================================================================================================
// Case Study: how ptgi::unique() works:
// Remember we "remove adjacent duplicates".
// In this example, the input is NOT fully sorted when ptgi:unique() is called.
//
// I put | separatators, BEFORE UNIQ to illustrate this
// 10 | 20,21,22 | 30,31 | 23,24 | 11
//
// In example above, 20, 21, 22 are "same" since dividing by 10 gives 2 quotient.
// And 30,31 are "same", since /10 quotient is 3.
// And 23, 24 are same, since /10 quotient is 2.
// And 11 is "group of one" by itself.
// So there are 5 groups, but the 4th group (23, 24) happens to be equal to group 2 (20, 21, 22)
// So there are 5 groups, and the 5th group (11) is equal to group 1 (10)
//
// R = result
// F = first
//
// 10, 20, 21, 22, 30, 31, 23, 24, 11
// R F
//
// 10 is result, and first points to 20, and R != F (10 != 20) so bump R:
// R
// F
//
// Now we hits the "optimized out swap logic".
// (avoid swap because R == F)
//
// // now bump F until R != F (integer division by 10)
// 10, 20, 21, 22, 30, 31, 23, 24, 11
// R F // 20 == 21 in 10x
// R F // 20 == 22 in 10x
// R F // 20 != 30, so we do a swap of ++R and F
// (Now first hits 21, 22, then finally 30, which is different than R, so we swap bump R to 21 and swap with 30)
// 10, 20, 30, 22, 21, 31, 23, 24, 11 // after R & F swap (21 and 30)
// R F
//
// 10, 20, 30, 22, 21, 31, 23, 24, 11
// R F // bump F to 31, but R and F are same (30 vs 31)
// R F // bump F to 23, R != F, so swap ++R with F
// 10, 20, 30, 22, 21, 31, 23, 24, 11
// R F // bump R to 22
// 10, 20, 30, 23, 21, 31, 22, 24, 11 // after the R & F swap (22 & 23 swap)
// R F // will swap 22 and 23
// R F // bump F to 24, but R and F are same in 10x
// R F // bump F, R != F, so swap ++R with F
// R F // R and F are diff, so swap ++R with F (21 and 11)
// 10, 20, 30, 23, 11, 31, 22, 24, 21
// R F // aftter swap of old 21 and 11
// R F // F now at last(), so loop terminates
// R F // bump R by 1 to point to dupPostion (first duplicate in range)
//
// return R which now points to 31
//==========================================================================================================
// NOTES:
// 1) the #ifdef IMPROVED_STD_UNIQUE_ALGORITHM documents how we have modified the original std::unique().
// 2) I've heavily unit tested this code, including using valgrind(1), and it is *believed* to be 100% defect-free.
//
//==========================================================================================================
// History:
// 130201 dpb dbednar@ptgi.com created
//==========================================================================================================
#ifndef PTGI_UNIQUE_HPP
#define PTGI_UNIQUE_HPP
// Created to solve memory leak problems when calling std::unique() on a vector<Route*>.
// Memory leaks discovered with valgrind and unitTesting.
#include <algorithm> // std::swap
// instead of std::myUnique, call this instead, where arg3 is a function ptr
//
// like std::unique, it puts the dups at the end, but it uses swapping to preserve original
// vector contents, to avoid memory leaks and duplicate pointers in vector<Object*>.
#ifdef IMPROVED_STD_UNIQUE_ALGORITHM
#error the #ifdef for IMPROVED_STD_UNIQUE_ALGORITHM was defined previously.. Something is wrong.
#endif
#undef IMPROVED_STD_UNIQUE_ALGORITHM
#define IMPROVED_STD_UNIQUE_ALGORITHM
// similar to std::unique, except that this version swaps elements, to avoid
// memory leaks, when vector contains pointers.
//
// Normally the input is sorted.
// Normal std::unique:
// 10 20 20 20 30 30 20 20 10
// a b c d e f g h i
//
// 10 20 30 20 10 | 30 20 20 10
// a b e g i f g h i
//
// Now GONE: c, d.
// Now DUPS: g, i.
// This causes memory leaks and segmenation faults due to duplicate deletes of same pointer!
namespace ptgi {
// Return the position of the first in range of duplicates moved to end of vector.
//
// uses operator== of class for comparison
//
// @param [first, last) is a range to find duplicates within.
//
// @return the dupPosition position, such that [dupPosition, end) are contiguous
// duplicate elements.
// IF all items are unique, then it would return last.
//
template <class ForwardIterator>
ForwardIterator unique( ForwardIterator first, ForwardIterator last)
{
// compare iterators, not values
if (first == last)
return last;
// remember the current item that we are looking at for uniqueness
ForwardIterator result = first;
// result is slow ptr where to store next unique item
// first is fast ptr which is looking at all elts
// the first iterator moves over all elements [begin+1, end).
// while the current item (result) is the same as all elts
// to the right, (first) keeps going, until you find a different
// element pointed to by *first. At that time, we swap them.
while (++first != last)
{
if (!(*result == *first))
{
#ifdef IMPROVED_STD_UNIQUE_ALGORITHM
// inc result, then swap *result and *first
// THIS IS WHAT WE WANT TO DO.
// BUT THIS COULD SWAP AN ELEMENT WITH ITSELF, UNCECESSARILY!!!
// std::swap( *first, *(++result));
// BUT avoid swapping with itself when both iterators are the same
++result;
if (result != first)
std::swap( *first, *result);
#else
// original code found in std::unique()
// copies unique down
*(++result) = *first;
#endif
}
}
return ++result;
}
template <class ForwardIterator, class BinaryPredicate>
ForwardIterator unique( ForwardIterator first, ForwardIterator last, BinaryPredicate pred)
{
if (first == last)
return last;
// remember the current item that we are looking at for uniqueness
ForwardIterator result = first;
while (++first != last)
{
if (!pred(*result,*first))
{
#ifdef IMPROVED_STD_UNIQUE_ALGORITHM
// inc result, then swap *result and *first
// THIS COULD SWAP WITH ITSELF UNCECESSARILY
// std::swap( *first, *(++result));
//
// BUT avoid swapping with itself when both iterators are the same
++result;
if (result != first)
std::swap( *first, *result);
#else
// original code found in std::unique()
// copies unique down
// causes memory leaks, and duplicate ptrs
// and uncessarily moves in place!
*(++result) = *first;
#endif
}
}
return ++result;
}
// from now on, the #define is no longer needed, so get rid of it
#undef IMPROVED_STD_UNIQUE_ALGORITHM
} // end ptgi:: namespace
#endif
这是我用来测试它的单元测试程序：
// QUESTION: in test2, I had trouble getting one line to compile,which was caused by the declaration of operator()
// in the equal_to Predicate. I'm not sure how to correctly resolve that issue.
// Look for //OUT lines
//
// Make sure that NOTES in ptgi_unique.hpp are correct, in how we should "cleanup" duplicates
// from both a vector<Integer> (test1()) and vector<Integer*> (test2).
// Run this with valgrind(1).
//
// In test2(), IF we use the call to std::unique(), we get this problem:
//
// [dbednar@ipeng8 TestSortRoutes]$ ./Main7
// TEST2: ORIG nums before UNIQUE: 10, 20, 21, 22, 30, 31, 23, 24, 11
// TEST2: modified nums AFTER UNIQUE: 10, 20, 30, 23, 11, 31, 23, 24, 11
// INFO: dupInx=5
// TEST2: uniq = 10
// TEST2: uniq = 20
// TEST2: uniq = 30
// TEST2: uniq = 33427744
// TEST2: uniq = 33427808
// Segmentation fault (core dumped)
//
// And if we run valgrind we seen various error about "read errors", "mismatched free", "definitely lost", etc.
//
// valgrind --leak-check=full ./Main7
// ==359== Memcheck, a memory error detector
// ==359== Command: ./Main7
// ==359== Invalid read of size 4
// ==359== Invalid free() / delete / delete[]
// ==359== HEAP SUMMARY:
// ==359== in use at exit: 8 bytes in 2 blocks
// ==359== LEAK SUMMARY:
// ==359== definitely lost: 8 bytes in 2 blocks
// But once we replace the call in test2() to use ptgi::unique(), all valgrind() error messages disappear.
//
// 130212 dpb dbednar@ptgi.com created
// =========================================================================================================
#include <iostream> // std::cout, std::cerr
#include <string>
#include <vector> // std::vector
#include <sstream> // std::ostringstream
#include <algorithm> // std::unique()
#include <functional> // std::equal_to(), std::binary_function()
#include <cassert> // assert() MACRO
#include "ptgi_unique.hpp" // ptgi::unique()
// Integer is small "wrapper class" around a primitive int.
// There is no SETTER, so Integer's are IMMUTABLE, just like in JAVA.
class Integer
{
private:
int num;
public:
// default CTOR: "Integer zero;"
// COMPRENSIVE CTOR: "Integer five(5);"
Integer( int num = 0 ) :
num(num)
{
}
// COPY CTOR
Integer( const Integer& rhs) :
num(rhs.num)
{
}
// assignment, operator=, needs nothing special... since all data members are primitives
// GETTER for 'num' data member
// GETTER' are *always* const
int getNum() const
{
return num;
}
// NO SETTER, because IMMUTABLE (similar to Java's Integer class)
// @return "num"
// NB: toString() should *always* be a const method
//
// NOTE: it is probably more efficient to call getNum() intead
// of toString() when printing a number:
//
// BETTER to do this:
// Integer five(5);
// std::cout << five.getNum() << "\n"
// than this:
// std::cout << five.toString() << "\n"
std::string toString() const
{
std::ostringstream oss;
oss << num;
return oss.str();
}
};
// convenience typedef's for iterating over std::vector<Integer>
typedef std::vector<Integer>::iterator IntegerVectorIterator;
typedef std::vector<Integer>::const_iterator ConstIntegerVectorIterator;
// convenience typedef's for iterating over std::vector<Integer*>
typedef std::vector<Integer*>::iterator IntegerStarVectorIterator;
typedef std::vector<Integer*>::const_iterator ConstIntegerStarVectorIterator;
// functor used for std::unique or ptgi::unique() on a std::vector<Integer>
// Two numbers equal if, when divided by 10 (integer division), the quotients are the same.
// Hence 50..59 are equal, 60..69 are equal, etc.
struct IntegerEqualByTen: public std::equal_to<Integer>
{
bool operator() (const Integer& arg1, const Integer& arg2) const
{
return ((arg1.getNum()/10) == (arg2.getNum()/10));
}
};
// functor used for std::unique or ptgi::unique on a std::vector<Integer*>
// Two numbers equal if, when divided by 10 (integer division), the quotients are the same.
// Hence 50..59 are equal, 60..69 are equal, etc.
struct IntegerEqualByTenPointer: public std::equal_to<Integer*>
{
// NB: the Integer*& looks funny to me!
// TECHNICAL PROBLEM ELSEWHERE so had to remove the & from *&
//OUT bool operator() (const Integer*& arg1, const Integer*& arg2) const
//
bool operator() (const Integer* arg1, const Integer* arg2) const
{
return ((arg1->getNum()/10) == (arg2->getNum()/10));
}
};
void test1();
void test2();
void printIntegerStarVector( const std::string& msg, const std::vector<Integer*>& nums );
int main()
{
test1();
test2();
return 0;
}
// test1() uses a vector<Object> (namely vector<Integer>), so there is no problem with memory loss
void test1()
{
int data[] = { 10, 20, 21, 22, 30, 31, 23, 24, 11};
// turn C array into C++ vector
std::vector<Integer> nums(data, data+9);
// arg3 is a functor
IntegerVectorIterator dupPosition = ptgi::unique( nums.begin(), nums.end(), IntegerEqualByTen() );
nums.erase(dupPosition, nums.end());
nums.erase(nums.begin(), dupPosition);
}
//==================================================================================
// test2() uses a vector<Integer*>, so after ptgi:unique(), we have to be careful in
// how we eliminate the duplicate Integer objects stored in the heap.
//==================================================================================
void test2()
{
int data[] = { 10, 20, 21, 22, 30, 31, 23, 24, 11};
// turn C array into C++ vector of Integer* pointers
std::vector<Integer*> nums;
// put data[] integers into equivalent Integer* objects in HEAP
for (int inx = 0; inx < 9; ++inx)
{
nums.push_back( new Integer(data[inx]) );
}
// print the vector<Integer*> to stdout
printIntegerStarVector( "TEST2: ORIG nums before UNIQUE", nums );
// arg3 is a functor
#if 1
// corrected version which fixes SEGMENTATION FAULT and all memory leaks reported by valgrind(1)
// I THINK we want to use new C++11 cbegin() and cend(),since the equal_to predicate is passed "Integer *&"
// DID NOT COMPILE
//OUT IntegerStarVectorIterator dupPosition = ptgi::unique( const_cast<ConstIntegerStarVectorIterator>(nums.begin()), const_cast<ConstIntegerStarVectorIterator>(nums.end()), IntegerEqualByTenPointer() );
// DID NOT COMPILE when equal_to predicate declared "Integer*& arg1, Integer*& arg2"
//OUT IntegerStarVectorIterator dupPosition = ptgi::unique( const_cast<nums::const_iterator>(nums.begin()), const_cast<nums::const_iterator>(nums.end()), IntegerEqualByTenPointer() );
// okay when equal_to predicate declared "Integer* arg1, Integer* arg2"
IntegerStarVectorIterator dupPosition = ptgi::unique(nums.begin(), nums.end(), IntegerEqualByTenPointer() );
#else
// BUGGY version that causes SEGMENTATION FAULT and valgrind(1) errors
IntegerStarVectorIterator dupPosition = std::unique( nums.begin(), nums.end(), IntegerEqualByTenPointer() );
#endif
printIntegerStarVector( "TEST2: modified nums AFTER UNIQUE", nums );
int dupInx = dupPosition - nums.begin();
std::cout << "INFO: dupInx=" << dupInx <<"\n";
// delete the dup Integer* objects in the [dupPosition, end] range
for (IntegerStarVectorIterator iter = dupPosition; iter != nums.end(); ++iter)
{
delete (*iter);
}
// shrink the vector
// NB: the Integer* ptrs are NOT followed by vector::erase()
nums.erase(dupPosition, nums.end());
// print the uniques, by following the iter to the Integer* pointer
for (IntegerStarVectorIterator iter = nums.begin(); iter != nums.end(); ++iter)
{
std::cout << "TEST2: uniq = " << (*iter)->getNum() << "\n";
}
// remove the unique objects from heap
for (IntegerStarVectorIterator iter = nums.begin(); iter != nums.end(); ++iter)
{
delete (*iter);
}
// shrink the vector
nums.erase(nums.begin(), nums.end());
// the vector should now be completely empty
assert( nums.size() == 0);
}
//@ print to stdout the string: "info_msg: num1, num2, .... numN\n"
void printIntegerStarVector( const std::string& msg, const std::vector<Integer*>& nums )
{
std::cout << msg << ": ";
int inx = 0;
ConstIntegerStarVectorIterator iter;
// use const iterator and const range!
// NB: cbegin() and cend() not supported until LATER (c++11)
for (iter = nums.begin(), inx = 0; iter != nums.end(); ++iter, ++inx)
{
// output a comma seperator *AFTER* first
if (inx > 0)
std::cout << ", ";
// call Integer::toString()
std::cout << (*iter)->getNum(); // send int to stdout
// std::cout << (*iter)->toString(); // also works, but is probably slower
}
// in conclusion, add newline
std::cout << "\n";
}
void removeDuplicates(std::vector<int>& arr) {
for (int i = 0; i < arr.size(); i++)
{
for (int j = i + 1; j < arr.size(); j++)
{
if (arr[i] > arr[j])
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
}
std::vector<int> y;
int x = arr[0];
int i = 0;
while (i < arr.size())
{
if (x != arr[i])
{
y.push_back(x);
x = arr[i];
}
i++;
if (i == arr.size())
y.push_back(arr[i - 1]);
}
arr = y;
}
如果您正在寻找性能和使用std::vector，我推荐这个文档链接提供的那个。
std::vector<int> myvector{10,20,20,20,30,30,20,20,10}; // 10 20 20 20 30 30 20 20 10
std::sort(myvector.begin(), myvector.end() );
const auto& it = std::unique (myvector.begin(), myvector.end()); // 10 20 30 ? ? ? ? ? ?
// ^
myvector.resize( std::distance(myvector.begin(),it) ); // 10 20 30
更易理解的代码来自：https ://en.cppreference.com/w/cpp/algorithm/unique
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cctype>
int main()
{
// remove duplicate elements
std::vector<int> v{1,2,3,1,2,3,3,4,5,4,5,6,7};
std::sort(v.begin(), v.end()); // 1 1 2 2 3 3 3 4 4 5 5 6 7
auto last = std::unique(v.begin(), v.end());
// v now holds {1 2 3 4 5 6 7 x x x x x x}, where 'x' is indeterminate
v.erase(last, v.end());
for (int i : v)
std::cout << i << " ";
std::cout << "\n";
}
输出：
1 2 3 4 5 6 7
std::set<int> s;
std::for_each(v.cbegin(), v.cend(), [&s](int val){s.insert(val);});
v.clear();
std::copy(s.cbegin(), s.cend(), v.cbegin());
关于 alexK7 基准测试。我尝试了它们并得到了类似的结果，但是当值的范围是 100 万时，使用 std::sort (f1) 和使用 std::unordered_set (f5) 的情况产生相似的时间。当值的范围是 1000 万时，f1 比 f5 快。
如果值的范围是有限的并且值是 unsigned int，则可以使用 std::vector，其大小对应于给定的范围。这是代码：
void DeleteDuplicates_vector_bool(std::vector<unsigned>& v, unsigned range_size)
{
std::vector<bool> v1(range_size);
for (auto& x: v)
{
v1[x] = true;
}
v.clear();
unsigned count = 0;
for (auto& x: v1)
{
if (x)
{
v.push_back(count);
}
++count;
}
}
如果您的类很容易转换为 int，并且您有一些内存，则可以在不进行排序的情况下完成唯一性，并且速度要快得多：
#include <vector>
#include <stdlib.h>
#include <algorithm>
int main (int argc, char* argv []) {
//vector init
std::vector<int> v (1000000, 0);
std::for_each (v.begin (), v.end (), [] (int& s) {s = rand () %1000;});
std::vector<int> v1 (v);
int beg (0), end (0), duration (0);
beg = clock ();
{
std::sort (v.begin (), v.end ());
auto i (v.begin ());
i = std::unique (v.begin (), v.end ());
if (i != v.end ()) v.erase (i, v.end ());
}
end = clock ();
duration = (int) (end - beg);
std::cout << "\tduration sort + unique == " << duration << std::endl;
int n (0);
duration = 0;
beg = clock ();
std::for_each (v1.begin (), v1.end (), [&n] (const int& s) {if (s >= n) n = s+1;});
std::vector<int> tab (n, 0);
{
auto i (v1.begin ());
std::for_each (v1.begin (), v1.end (), [&i, &tab] (const int& s) {
if (!tab [s]) {
*i++ = s;
++tab [s];
}
});
std::sort (v1.begin (), i);
v1.erase (i, v1.end ());
}
end = clock ();
duration = (int) (end - beg);
std::cout << "\tduration unique + sort == " << duration << std::endl;
if (v == v1) {
std::cout << "and results are same" << std::endl;
}
else {
std::cout << "but result differs" << std::endl;
}
}
典型结果：duration sort + unique == 38985 duration unique + sort == 2500 并且结果相同
大多数答案似乎都在使用O(nlogn)，但使用unordered_set我们可以将其减少到O(n). 我看到了一些使用 的解决方案，但我发现了这个，使用andsets似乎更优雅。setiterators
using Intvec = std::vector<int>;
void remove(Intvec &v) {
// creating iterator starting with beginning of the vector
Intvec::iterator itr = v.begin();
std::unordered_set<int> s;
// loops from the beginning to the end of the list
for (auto curr = v.begin(); curr != v.end(); ++curr) {
if (s.insert(*curr).second) { // if the 0 curr already exist in the set
*itr++ = *curr; // adding a position to the iterator
}
}
// erasing repeating positions in the set
v.erase(itr, v.end());
}
这是 std::unique() 发生的重复删除问题的示例。在 LINUX 机器上，程序崩溃。阅读评论了解详情。
// Main10.cpp
//
// Illustration of duplicate delete and memory leak in a vector<int*> after calling std::unique.
// On a LINUX machine, it crashes the progam because of the duplicate delete.
//
// INPUT : {1, 2, 2, 3}
// OUTPUT: {1, 2, 3, 3}
//
// The two 3's are actually pointers to the same 3 integer in the HEAP, which is BAD
// because if you delete both int* pointers, you are deleting the same memory
// location twice.
//
//
// Never mind the fact that we ignore the "dupPosition" returned by std::unique(),
// but in any sensible program that "cleans up after istelf" you want to call deletex
// on all int* poitners to avoid memory leaks.
//
//
// NOW IF you replace std::unique() with ptgi::unique(), all of the the problems disappear.
// Why? Because ptgi:unique merely reshuffles the data:
// OUTPUT: {1, 2, 3, 2}
// The ptgi:unique has swapped the last two elements, so all of the original elements in
// the INPUT are STILL in the OUTPUT.
//
// 130215 dbednar@ptgi.com
//============================================================================
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
#include "ptgi_unique.hpp"
// functor used by std::unique to remove adjacent elts from vector<int*>
struct EqualToVectorOfIntegerStar: public std::equal_to<int *>
{
bool operator() (const int* arg1, const int* arg2) const
{
return (*arg1 == *arg2);
}
};
void printVector( const std::string& msg, const std::vector<int*>& vnums);
int main()
{
int inums [] = { 1, 2, 2, 3 };
std::vector<int*> vnums;
// convert C array into vector of pointers to integers
for (size_t inx = 0; inx < 4; ++ inx)
vnums.push_back( new int(inums[inx]) );
printVector("BEFORE UNIQ", vnums);
// INPUT : 1, 2A, 2B, 3
std::unique( vnums.begin(), vnums.end(), EqualToVectorOfIntegerStar() );
// OUTPUT: 1, 2A, 3, 3 }
printVector("AFTER UNIQ", vnums);
// now we delete 3 twice, and we have a memory leak because 2B is not deleted.
for (size_t inx = 0; inx < vnums.size(); ++inx)
{
delete(vnums[inx]);
}
}
// print a line of the form "msg: 1,2,3,..,5,6,7\n", where 1..7 are the numbers in vnums vector
// PS: you may pass "hello world" (const char *) because of implicit (automatic) conversion
// from "const char *" to std::string conversion.
void printVector( const std::string& msg, const std::vector<int*>& vnums)
{
std::cout << msg << ": ";
for (size_t inx = 0; inx < vnums.size(); ++inx)
{
// insert comma separator before current elt, but ONLY after first elt
if (inx > 0)
std::cout << ",";
std::cout << *vnums[inx];
}
std::cout << "\n";
}
void EraseVectorRepeats(vector <int> & v){
TOP:for(int y=0; y<v.size();++y){
for(int z=0; z<v.size();++z){
if(y==z){ //This if statement makes sure the number that it is on is not erased-just skipped-in order to keep only one copy of a repeated number
continue;}
if(v[y]==v[z]){
v.erase(v.begin()+z); //whenever a number is erased the function goes back to start of the first loop because the size of the vector changes
goto TOP;}}}}
这是我创建的一个函数，您可以使用它来删除重复。所需的头文件只是<iostream>和<vector>.