From 6a4947d83da54ab58a39c5295c51f85e3da15921 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, 15 Sep 2018 23:01:38 +0800
Subject: [PATCH] Create 389. Find the Difference.md

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 389. Find the Difference.md | 22 ++++++++++++++++++++++
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 create mode 100644 389. Find the Difference.md

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+# [389. Find the Difference](https://leetcode.com/problems/find-the-difference/description/)
+# 思路
+分别记录两个字符串中每个字符的出现次数，最后比较一下字符的出现次数，次数差1对应的字符即所求。   
+因为全是小写字母，所以开辟一个大小为26的数组进行计数即可。另外，没必要开辟两个计数数组，用一个数组记录即可，对s中出现的字符进行次数累加，对t中出现的字符进行次数累减。
+最后次数为-1的对应的字母即所求。   
+时间复杂度O(n)， 空间复杂度O(1)
+# C++
+```
+class Solution {
+public:
+    char findTheDifference(string s, string t) {
+        vector<int>count(26, 0);
+        for(int i = 0; i < s.size(); i++){
+            count[s[i] - 'a']++;
+            count[t[i] - 'a']--;
+        }
+        count[t[t.size() - 1] - 'a']--;
+        for(int i = 0; i < 26; i++)
+            if(count[i] == -1) return char(i + 'a');
+    }
+};
+```