# [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/description/)
# 思路
从前往后遍历一遍，用currsum记录以当前位置为结尾的最大子序列和。可见currsum要么等于nums[i]本身，要么等于nums[i]加上上一个currsum，即更新准则为：  
如果currsum大于0，那么currsum += nums[i]；  
否则currsum = nums[i]  
# C++
```
class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int maxsum = nums[0]; // at least one number
        int currsum = nums[0];
        for(int i = 1; i < nums.size(); i++){
            if(currsum > 0) currsum += nums[i];
            else currsum = nums[i];
            
            if(currsum > maxsum) maxsum = currsum;
        }
        return maxsum;
    }
};
```