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solutions/53. Maximum Subarray.md
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# [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/description/) # [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/description/)
# 思路 # 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： 就是一个简单的动态规划。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]； * 如果currsum大于0，那么`currsum += nums[i]`；
否则currsum = nums[i] * 否则，则`currsum = nums[i]`。
时间复杂度O(n)，空间复杂度O(1)
# C++ # C++
``` C++ ``` C++
class Solution { class Solution {