第三周作业1(LeetCode53)

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1. 题目描述(LeetCode53)

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

2. 解决思路

利用线性时间求解，时间复杂度为O(n)。

3. 完整代码

#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>#include <stdlib.h>void MaxSubSum(int *arr, int len, int &l, int &r);int main(){    int* arr = NULL;  //存储数组    int len = 0;    printf("请输入数组的长度：\n");    scanf("%d", &len);    while (len <= 0)    {        printf("长度必须大于0，请重新输入：\n");        scanf("%d", &len);    }    arr = (int*)malloc(sizeof(int)* len);  //数组分配内存    printf("请输入数组的每一位数：\n");    for (int i = 0; i<len; ++i)    {        scanf("%d", &arr[i]);    }    //调用封装好的功能函数    int l, r;    MaxSubSum(arr, len, l, r);    //输出结果    printf("新的数组长度为: %d\n", r - l + 1);    printf("和最大的连续子数组元素为:\n");    for (int i = l; i <= r; ++i)    {        printf("%d ", arr[i]);    }    scanf("%d", &len);  //加个输入让窗口停下来    free(arr);  //释放内存    return 0;}void MaxSubSum(int *arr, int len, int &l, int &r){    int i;    int MaxSum = 0;    int CurSum = 0;    int ll = -1;    for (i = 0; i < len; i++)    {        CurSum += arr[i];        if (CurSum > MaxSum)        {            MaxSum = CurSum;            l = ll + 1;            r = i;        }        //如果累加和出现小于0的情况        //则和最大的子序列肯定不可能包含前面的元素        //这时累加和置零，从下个元素重新开始累加        if (CurSum <= 0)        {            CurSum = 0;            ll = i;        }    }}

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