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.
click to show more practice.
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.
思路 ：
连续子数组最大和问题，当以第i-1 个数字结尾的子数组的和，小于0时，如果把这个负数与第i 个数相加，得到的结果第i 个数本身还要小，所以此时不能把它加到第i 个数上。
sum条件sum = nums[i]如果当前的sum 小于0sum = sum + nums[i]如果当前的sum 大于0 class Solution { public int maxSubArray(int[] nums) { int max = nums[0]; int sum = nums[0]; for(int i = 1; i < nums.length; i++) { if(sum <= 0) { sum = nums[i]; }else { sum += nums[i]; } max = Math.max(max, sum); } return max; } }注：学渣心里苦，不要学楼主，平时不努力，考试二百五，哭 ~
