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;
    }
}