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子数组包含正数和负数。您必须找到一个最大和子数组，使得子数组的长度大于或等于 k。

这是我使用 Kadane 算法的 c++ 代码。

#include <iostream>

using namespace std;

int main(){

    int n,k;
    cin >> n >> k;
    int array[n];
    int sum = 0;
    int maxsum = 0;
    int beststarts[n];

    for(int i = 0;i < n; i++){
            cin >> array[i];
    }

    for(int i = 0;i < k-1;i ++){
            sum = sum+array[i];
            beststarts[i] = 0;
    }


    for(int i =  k-1;i < n; i++){ //best end search with min length;
            sum = sum+array[i];
            int testsum = sum;
            if(i > 0){
            beststarts[i] = beststarts[i-1];
            }
            for(int j = beststarts[i] ;i-j > k-1;j ++){
                    testsum = testsum - array[j];
                    if(testsum > sum){
                            beststarts[i] = j+1;
                            sum = testsum;
                    }
            }
            if(sum > maxsum){
                    maxsum = sum;
            }
    }

    cout << maxsum;

    return 0;
}

我的代码运行良好，但速度很慢，我想不出任何方法来改进我的代码。我也读过这个问题查找总和可被 K 整除的最长子数组，但这不是我想要的，长度也可以大于 k。

4

2 回答 2

3

基于此答案的解决方案

现场演示

#include <algorithm>
#include <iterator>
#include <iostream>
#include <numeric>
#include <ostream>
#include <utility>
#include <vector>

// __________________________________________________

template<typename RandomAccessIterator> typename std::iterator_traits<RandomAccessIterator>::value_type
max_subarr_k(RandomAccessIterator first,RandomAccessIterator last,int k)
{
    using namespace std;
    typedef typename iterator_traits<RandomAccessIterator>::value_type value_type;
    if(distance(first,last) < k)
        return value_type(0);
    RandomAccessIterator tail=first;
    first+=k;
    value_type window=accumulate(tail,first,value_type(0));
    value_type max_sum=window, current_sum=window;
    while(first!=last)
    {
        window += (*first)-(*tail) ;
        current_sum = max( current_sum+(*first), window );
        max_sum = max(max_sum,current_sum);
        ++first;
        ++tail;
    }
    return max_sum;
}

// __________________________________________________

template<typename E,int N>
E *end(E (&arr)[N])
{
    return arr+N;
}

int main()
{
    using namespace std;
    int arr[]={1,2,4,-5,-4,-3,2,1,5,6,-20,1,1,1,1,1};
    cout << max_subarr_k(arr,end(arr),4) << endl;
    cout << max_subarr_k(arr,end(arr),5) << endl;
}

输出是：

14
11
于 2012-11-02T05:33:18.690 回答
0 
  int w(0);
    for (int i=0; i < k; i++) w += a[i];
    int run_sum(w), max_sum(w);
    for (int i=k; i < n; i++) {
              w = a[i] + max(w, w-a[i-k]); //  window will such that it will include run_sum
              run_sum = max(run_sum + a[i], w);
              max_sum = max(run_sum, max_sum); 
    }
    return max_sum;
于 2013-07-29T11:23:28.107 回答