From 38de2caff8c71b7a8e358c80b7fb75cdb2cc2559 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Sat, 1 Sep 2018 19:45:24 +0800
Subject: [PATCH] Create 53. Maximum Subarray.md

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 53. Maximum Subarray.md | 22 ++++++++++++++++++++++
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 create mode 100644 53. Maximum Subarray.md

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+# [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;
+    }
+};
+```