求子數組的最大和

輸入一個整形數組，數組裏有正數也有負數。
數組中連續的一個或多個整數組成一個子數組，每一個子數組都有一個和。
求全部子數組的和的最大值。要求時間複雜度爲O(n)。

例如輸入的數組爲1, -2, 3, 10, -4, 7, 2, -5，和最大的子數組爲3, 10, -4, 7, 2，
所以輸出爲該子數組的和18。

#include<stdio.h>
int subStrSum(int* s, int len)
{
    int i = 1, curSum = s[0], maxSum = s[0];
    while(i < len)
    {
        if(curSum > 0 &&curSum + s[i] > 0)//當前和爲負數或者加當前元素後爲負數，則不進行累加，而賦當前和爲當前元素
            curSum += s[i];
        else
            curSum = s[i];
        i += 1;
        if(curSum > maxSum)
            maxSum = curSum;
    }
    return maxSum;
}
void main(){
    int s[] = {-2,-2,-5,-2,-1,-6,-11,-7,-3};
    int maxSum = subStrSum(s,9);
    printf("%d\n",maxSum);
}

 改進，輸出該子數組

#include<stdio.h>
typedef struct subStr{
    int start;
    int length;
    int sum;
}str;
str subStrSum(int* s, int len)
{
    str s1;
    int i = 1, curSum = s[0];
    s1.start = 0;
    s1.length =1;
    s1.sum = s[0];
    int length = 1, start = 0;
    while(i < len)
    {
        if(curSum > 0 &&curSum + s[i] > 0)
        {
            curSum += s[i];
            length += 1;
        }
        else
        {
            curSum = s[i];
            start = i;
            length = 1;
        }      
        if(curSum > s1.sum)
        {
            s1.start = start;
            s1.length = length;
            s1.sum = curSum;
        }
        i += 1;
    }
    return s1;
}
void main(){
    int s[] = {1,2,-5,2,1,6,-11,7,-3};
    str str1 = subStrSum(s,9);
    int i ;
    for(i = 0; i < str1.length; i++)
        printf("%d,",s[str1.start+i]);
    printf("sum is %d\n",str1.sum);
}