From cb30e5543e2cd9e931a56ddfb331b581d9e4156d 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: Wed, 12 Sep 2018 22:06:36 +0800
Subject: [PATCH] Create 136. Single Number.md

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+# [136. Single Number](https://leetcode.com/problems/single-number/description/)
+# 思路
+## 思路一 
+最常规的思路就是用一个map或者散列(unordered_map)记录是否出现过。   
+map: 时间复杂度O(nlogn), 空间复杂度O(n)   
+unordered_map: 时间复杂度O(n), 空间复杂度O(N)
+
+## 思路二*
+若某个数出现两次，则异或操作后得0，所以可以考虑将数组所有元素进行异或操作，最终得到的值就是欲求值。   
+时间复杂度O(n), 空间复杂度O(1), 完美
+# C++
+## 思路二
+```
+class Solution {
+public:
+    int singleNumber(vector<int>& nums) {
+        int res = 0;
+        for(int num: nums) res = res ^ num;
+        
+        return res;
+    }
+};
+```