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?
思路：a^b^b=a a抑或同一个数两次认为a。
1 class Solution { 2 public: 3 int singleNumber(int A[], int n) { 4 int res; 5 for(int i=0;i<n;i++) 6 { 7 res=res^A[i]; 8 } 9 return res; 10 } 11 };
原文：http://www.cnblogs.com/hicandyman/p/3825802.html