53. Maximum Subarray

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博客围绕整数数组展开，要找出其中具有最大和的连续子数组并返回其和。给出了遍历数组的求解方法，用变量记录当前数字前的和最大值，根据与当前数相加情况更新该变量。还提到可用分治法求解。

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

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

遍历一次数组，用一个变量记录当前数字之前的和最大值，如果加上当前数更大，则更新，如果当前数更大，则此变量为当前数。

 class Solution {
        public int maxSubArray(int[] nums) {
            int ans = Integer.MIN_VALUE, sum = 0;
            for (int num : nums) {
                //加上num更大或者num更大，即如果前面数字为负数，则更新sum为num
                sum = Math.max(sum + num, num);
                ans = Math.max(ans, sum);
            }
            return ans;
        }
}