求和最大的连续子串

思路：

采用动态规划进行求解，令：
maxSum[k]表示子串list{0,k}的最大连续子串和,
maxSumIncludeK[k]表示子串list{0,k}中末尾为k的最大连续子串和，
则
maxSumIncludeK[k] = max(list[k],list[k]+maxSumIncludeK[k-1]),
maxSum[k] = max(maxSumIncludeK[k],maxSum[k-1])。

代码：

#include <stdlib.h>
#include <stdio.h> 

int MaxSubList(int list[],int length)
{
    int i;
    //maxSum[k]表示子串list{0,k}的最大连续子串和
    //maxSumIncludeK[k]表示子串list{0,k}中末尾为k的最大连续子串和
    int* maxSum = (int*)malloc(sizeof(int)*length);
    int* maxSumIncludeK = (int*)malloc(sizeof(int)*length);
    for (i=0;i<length;i++)      
    {          
        maxSum[i] = 0;          
        maxSumIncludeK[i] = 0;      
    }      
    if (list[0]>0) maxSum[0] = list[0];
    maxSumIncludeK[0] = list[0];
    for (i=1;i<length;i++)      
    {          
        maxSumIncludeK[i] = list[i];          
        if (maxSumIncludeK[i-1]>0) maxSumIncludeK[i]+=maxSumIncludeK[i-1];
        if (maxSumIncludeK[i]>maxSum[i-1]) maxSum[i] = maxSumIncludeK[i];
        else maxSum[i] = maxSum[i-1];
    }
    return maxSum[length-1];
} 

int main()
{
    int list[20] = {4,9,-4,3,2,-1,-6,7,2,-3,1,5,3,-2,7,-3,9,2,-4,2};
    printf("max sum:%d \n",MaxSubList(list,20));
    return 0;
}