标签：lin int cep number turn ons cto run lex
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?
class Solution { public: int singleNumber(vector<int>& nums) { int ans = 0; vector<int>::const_iterator it = nums.cbegin(); for (; it != nums.cend(); ++it) { ans^=*it; } return ans; } };
标签：lin int cep number turn ons cto run lex
原文地址：http://www.cnblogs.com/pk28/p/7230136.html
