Question

#include <iostream>
#include <limits>
int MIN = std::numeric_limits<int>::min()
using namespace std ; 

void findMaxSubArray(int inputArray[] , int n )
{

int maxStartIndex=0;
int maxEndIndex=0;
int maxSum = MIN ; 

int cumulativeSum= 0;
int maxStartIndexUntilNow=0;

for (int currentIndex = 0; currentIndex < n ; currentIndex++) 
{

    int eachArrayItem = inputArray[currentIndex];

    cumulativeSum+=eachArrayItem;

    if(cumulativeSum>maxSum)
    {
        maxSum = cumulativeSum;
        maxStartIndex=maxStartIndexUntilNow;
        maxEndIndex = currentIndex;
    }
    else if (cumulativeSum<0)
    {
        maxStartIndexUntilNow=currentIndex+1;
        cumulativeSum=0;
    }
}

cout << "Max sum         : "<< maxSum << "\n" ;
cout << "Max start index : "<< maxStartIndex << "\n" ;
cout << "Max end index   : "<< maxEndIndex << "\n" ;
}

int main() 
{
    int intArr[] = {-1,3,-1,-1,-1,-1,-1,-1 } ;
    //int intArr[] = {-1, 3, -5, 4, 6, -1, 2, -7, 13, -3};
    //int intArr[]={-6,-2,-3,-4,-1,-5,-5};
    findMaxSubArray(intArr,8);
    return 0 ; 
}

我对给定here的实现是否正确持怀疑态度，因此我在C ++中完全实现了它，而对于上述测试用例，它并不起作用。我无法找出算法出错的地方？

Answer 1

取 int maxSum = -1;可以解决您的问题。你的上述程序也不可编译。这适用于integer个数字

#include <iostream>
#include <limits>
using namespace std ; 

int MIN = std::numeric_limits<int>::min();
void findMaxSubArray(int inputArray[] , int n )
{

    int maxStartIndex=0;
    int maxEndIndex=0;
    int maxSum = -1 ; 

    int cumulativeSum= 0;
    int maxStartIndexUntilNow=0;

    for (int currentIndex = 0; currentIndex < n ; currentIndex++) 
    {

        int eachArrayItem = inputArray[currentIndex];

        cumulativeSum+=eachArrayItem;

        if(cumulativeSum>maxSum)
        {
            maxSum = cumulativeSum;
            maxStartIndex=maxStartIndexUntilNow;
            maxEndIndex = currentIndex;
        }
        else if (cumulativeSum<0)
        {
            maxStartIndexUntilNow=currentIndex+1;
            cumulativeSum=0;
        }
    }

    cout<< "Max sum         : "<< maxSum << "\n" ;
    cout<< "Max start index : "<< maxStartIndex << "\n" ;
    cout<< "Max end index   : "<< maxEndIndex << "\n" ;
}

int main() 
{
    int intArr[] = {-1,3,-1,-1,-1,-1,-1,-1 } ;
    //int intArr[] = {-1, 3, -5, 4, 6, -1, 2, -7, 13, -3};
    //int intArr[]={-6,-2,-3,-4,-1,-5,-5};
    findMaxSubArray(intArr,8);
    return 0 ; 
}

Answer 2

int maxStartIndex=0;
int maxEndIndex=0;
int maxSum = MIN;

这是你的问题。你在骗算。在索引0处开始和结束的子数组的总和为arr[0]，而不是负无穷大。但这也不是一个好的起点。

int maxStartIndex=0;
int maxEndIndex=-1;
int maxSum = 0;

任何数组都有一个零和子数组：一个空数组。你需要击败那个，而不是任何负数。

Answer 3

总的来说，那里有很多好的资源。这是a useful resource that you should look at for C++的链接。您还可以查看this resource which is where the code below originates from and which has a C implementation。这是粗略算法的伪代码：

Initialize:
    max_so_far = 0
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_ending_here < 0)
            max_ending_here = 0
  (c) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
return max_so_far

这是一个在C中实现算法的简单程序：

#include<stdio.h>
int maxSubArraySum(int a[], int size)
{
   int max_so_far = 0, max_ending_here = 0;
   int i;
   for(i = 0; i < size; i++)
   {
     max_ending_here = max_ending_here + a[i];
     if(max_ending_here < 0)
        max_ending_here = 0;
     if(max_so_far < max_ending_here)
        max_so_far = max_ending_here;
    }
    return max_so_far;
} 

/*Driver program to test maxSubArraySum*/
int main()
{
   int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
   int n = sizeof(a)/sizeof(a[0]);
   int max_sum = maxSubArraySum(a, n);
   printf("Maximum contiguous sum is %d\n", max_sum);
   getchar();
   return 0;
}

正如人们所指出的，这种方法并不适用于所有负数。如果所有数字都是负数，它只返回0。因此，我们可以进一步优化问题。下面是一些示例代码，可以很好地优化原始方法：

int maxSubArraySum(int a[], int size)
{
   int max_so_far = 0, max_ending_here = 0;
   int i;
   for(i = 0; i < size; i++)
   {
     max_ending_here = max_ending_here + a[i];
     if(max_ending_here < 0)
         max_ending_here = 0;

     /* Do not compare for all elements. Compare only   
        when  max_ending_here > 0 */
     else if (max_so_far < max_ending_here)
         max_so_far = max_ending_here;
   }
   return max_so_far;
}

Answer 4

您的代码存在的问题是（cumulativeSum＆gt; maxSum）在此之前检查（累积＆lt; 0）并且您的maxSum是MIN，因此如果第一个数字是负数而第二个数字是正数，则它将作为cumulativeSum＆gt;失败。 maxSum so -1将被添加到cumulativeSum，因此答案将是2而不是3.因此要么检查（cumulativeSum＆lt; 0）之前或者使maxSum = -1或添加条件（cumulativeSum＆gt; maxSum＆amp;＆amp; cumulativeSum＆gt; 0 ）

这是Kadane算法的错误实现吗？

4 个答案: