题目：
Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
代码：
class Solution { public: int singleNumber(vector<int>& nums) { int result = 0; const int W = sizeof(int)*8; int *count = new int[W](); for (size_t i = 0; i < nums.size(); ++i) { for (size_t j = 0; j < W; ++j) { count[j] += (nums[i] >> j) & 1; } } for (size_t i = 0; i < W; ++i) { if ( count[i]%3!=0 ) { result += ( 1 << i); } } delete []count; return result; } };
Tips:
1. 算法原理：由于都是int型的，只需要记录每个bit上1出现的次数；如果1出现的次数不是3的倍数，则将该位置置为1，并赋值到result中。
这道题主要是熟悉下位运算的各种操作，有时巧用位运算，可以大大提升代码效率。
原文：http://www.cnblogs.com/xbf9xbf/p/4461286.html