LeetCode Problem 136:Single Number

描述：Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

题目要求O(n)时间复杂度，O(1)空间复杂度。

思路1：初步使用暴力搜索，遍历数组，发现两个元素相等，则将这两个元素的标志位置为1，最后返回标志位为0的元素即可。时间复杂度O(n^2)没有AC，Status:Time Limit Exceed

 class Solution {
 public:
     int singleNumber(int A[], int n) {

         vector <int> flag(n,);

         for(int i = ; i < n; i++)  {
             if(flag[i] == )
                 continue;
             else    {
                 for(int j = i + ; j < n; j++)  {
                     if(A[i] == A[j])    {
                         flag[i] = ;
                         flag[j] = ;
                     }
                 }
             }
         }

         for(int i = ; i < n; i++)  {
             if(flag[i] == )
                 return A[i];
         }
     }
 };

思路2：利用异或操作。异或的性质1：交换律a ^ b = b ^ a，性质2：0 ^ a = a。于是利用交换律可以将数组假想成相同元素全部相邻，于是将所有元素依次做异或操作，相同元素异或为0，最终剩下的元素就为Single Number。时间复杂度O(n)，空间复杂度O(1)

 class Solution {
 public:
     int singleNumber(int A[], int n) {

         //异或
         int elem = ;
         for(int i = ; i < n ; i++) {
             elem = elem ^ A[i];
         }

         return elem;
     }
 };