LeetCode OJ平台上Maximum Subarray题目O(n)复杂度解决方案

原始题目如下，意为寻找数组和最大的子串，返回这个最大和即可。

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.

最普通的方法就是两层循环来寻找，复杂度为O(n^2).

在木易先森的指导下，有一个极其简单的O(n)复杂度的方法：

- 先找出数组max值，如果max小于0 ，啥也别说了，直接返回max.

- 设置辅助变量currentmax=0;数组从头至尾扫描一遍

- 如果currentmax + A[i] <= 0，意味着这个子串对于我们寻找最大和是没有任何帮助的，此时，直接再置currentmax = 0

- 如果currentmax + A[i] > 0，意味着这个子串的和是有意义的，将这个和跟max比较，时刻保持max的值是当前最理想的最大值

- 最后返回max即可

源代码如下：

public class Solution {public static int maxSubArray(int[] A) {if(A.length == 0)return 0;int max = A[0];for(int i = 0; i < A.length; i ++)if(A[i] > max)max = A[i];if(max <= 0)return max;int currentmax = 0;for(int i = 0;i < A.length; i ++){currentmax += A[i];if(currentmax <= 0){currentmax = 0;continue;}else{if(currentmax > max)max = currentmax;}}return max;}
}

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