Maximum Subarray

题目描述：
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.

想到了二分法。
这个最大值要么包括nums[mid]，要么就是在mid的左边或者右边。
要考虑到细节，不然就会出错！代码如下：

public class Solution {    public int maxSubArray(int[] nums) {        return findMaxSub(nums, 0, nums.length-1);    }public int findMaxSub(int[] nums,int left,int right){if(left==right)return nums[left];if(left>right)return Integer.MIN_VALUE;int mid=(left+right)/2,midSum=nums[mid];int midLeft=mid-1,midLeftsum=0,leftSum=Integer.MIN_VALUE;while(midLeft>=left){midLeftsum+=nums[midLeft--];if(midLeftsum>leftSum)leftSum=midLeftsum;}int midRight=mid+1,midRightSum=0,rightSum=Integer.MIN_VALUE;while(midRight<=right){midRightSum+=nums[midRight++];if(midRightSum>rightSum){rightSum=midRightSum;}}if(leftSum>0)midSum+=leftSum;if(rightSum>0)midSum+=rightSum;int leftMaxSub=findMaxSub(nums, left, mid-1);int rightMaxSub=findMaxSub(nums, mid+1, right);return Integer.max(Integer.max(leftMaxSub, rightMaxSub),midSum);}}

这个题用动态规划解是最简单的：

public class Solution {    public int maxSubArray(int[] A) {        int[] sum = new int[A.length];                int max = A[0];        sum[0] = A[0];        for (int i = 1; i < A.length; i++) {            sum[i] = Math.max(A[i], sum[i - 1] + A[i]);            max = Math.max(max, sum[i]);        }        return max;    }}