求最长子序列和

法一：O（n^3）

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

int sub_Sum(int a[], int left, int right)
{
    int i, temp_sum = 0;
    for(i = left; i <= right; i++)
    {
        temp_sum += a[i];
    }
    return temp_sum;
}

int main()
{
    int i, j, k, len;
    int max_sum = 0, cur_sum = 0;
    printf("Please input the length of the sequence: \n");
    scanf("%d\n", &len);
    int *a = (int *)malloc(len * sizeof(int));

    for(i = 0; i < len; i++)
    {
        scanf("%d", &a[i]);
    }

    for(i = 0; i < len; i++)
    {
        for(j = i; j < len; j++)
        {
            for(k = i; k <= j; k++)
            {
                cur_sum += a[k];
            }
            if(cur_sum > max_sum)
                max_sum = cur_sum;
            cur_sum = 0;
        }
    }
    printf("Max sum is : %d\n", max_sum);

   max_sum = 0;
    for(i = 0; i < len; i++)
    {
        for(j = i; j < len; j++)
        {
            cur_sum = sub_Sum(a, i, j);
            if(cur_sum > max_sum)
                max_sum = cur_sum;
            cur_sum = 0;
        }
    }
    printf("Max sum is : %d\n", max_sum);

    free(a);
    return 0;
}

法二：O(n^2)

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

int main()
{
    int i, j, len;
    int cur_sum = 0, max_sum = 0;
    printf("Please input the length of the sequence : \n");
    scanf("%d\n", &len);
    int *a = (int *)malloc(len * sizeof(int));
    for(i = 0; i < len; i++)
    {   
        scanf("%d", &a[i]);
    }   
    for(i = 0; i < len; i++)
    {   
        //计算一次以i开始的最大和值
        for(j = i; j < len; j++)
        {   
            cur_sum += a[j];
            if(cur_sum > max_sum)
            {   
                max_sum = cur_sum;
            }   
        }   
        cur_sum = 0;
    }   
    printf("The max sum is: %d\n", max_sum);
    free(a);

    return 0;
}

法三：分治法O(nlogn)

分治的思想：最大子序列和可能出现在三处。或者整个出现在输入数据的左半部，或者整个出现右半部，或者跨越输入数据的中部从而占据左右两个半部分。前两种情况递归求解。第三种情况的最大和可以通过求出前半部分的最大和（包含前半部分的最后一个元素）以及后半部分的最大和（包含后半部分的第一个元素)而得到，然后将这两个和加在一起，求出三个值的最大值。

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

int DivideAndConque_sum(int a[], int left, int right)
{
    int i, j;
    if(left == right)
    {
        if(a[left] > 0)
            return a[left];
        else
            return 0;
    }
    int center = (left + right) / 2;
    //每次递归返回时，该值为该字段的最终左最大子序列的和
    int maxLeftSum = DivideAndConque_sum(a, left, center);
    //每次递归返回时，该值为该字段的右最大子序列和
    int maxRightSum = DivideAndConque_sum(a, center + 1, right);

    //从中间向左扩展求子段的最大子序列和
    int maxLeftBorderSum = 0, leftBorderSum = 0;
    for(i = center; i >= left; i--)
    {
        leftBorderSum += a[i];
        if(leftBorderSum > maxLeftBorderSum)
            maxLeftBorderSum = leftBorderSum;
    }

    //从中间向右扩展求子段的最大子序列和
    int maxRightBorderSum = 0, rightBorderSum = 0;
    for(j = center + 1; j < right; j++)
    {
        rightBorderSum += a[j];
        if(rightBorderSum > maxRightBorderSum)
            maxRightBorderSum = rightBorderSum;
    }

    int tmp = maxLeftSum > maxRightSum ? maxLeftSum : maxRightSum;
    if(tmp >= maxLeftBorderSum + maxRightBorderSum)
        return tmp;
    else
        return maxLeftBorderSum + maxRightBorderSum;
}
int main()
{
    int i, len;
    printf("Please input the length of the sequence : \n");
    scanf("%d", &len);
    int *a = (int *)malloc(len * sizeof(int));
    for(i = 0; i < len; i++)
    {
        scanf("%d", &a[i]);
    }
    printf("The max sub length is : %d\n", DivideAndConque_sum(a, 0, len));
    return 0;
}

法四：动态规划O(n)

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

int main()
{
    int i, len;
    int curSum = 0, maxSum = 0;
    printf("Input the length of the sequence: \n");
    scanf("%d", &len);
    int *a = (int *)malloc(len * sizeof(int));
    for(i = 0; i < len; i++)
    {   
        scanf("%d", &a[i]);
    }   
    for(i = 0; i < len; i++)
    {   
        curSum = curSum + a[i];

        if(curSum > maxSum)
        {   
            maxSum = curSum;
        }   

        if(curSum < 0)
            curSum = 0;
    }   
    printf("The max length of the sequence is : %d\n", maxSum);
}