题目描述：
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?
解题方案：
该题目的一般方法是开辟一个空间为32的数组bitsum[32]，然后遍历题目中给定的数组，将数组中每个整数的每一bit位给分出来，加到bitsum的对应位置。最后再遍历bitsum数组，同时求bitsum[i]%3，将结果做移位操作！下面是该题的代码：
1 class Solution { 2 public: 3 int singleNumber(int A[], int n) { 4 int bitsum[32] = {0}; 5 int result = 0; 6 for (int i = 0; i < n; ++i) { 7 for (int j = 0; j < 32; ++j) { 8 bitsum[j] += A[i]>>j & 1; 9 } 10 } 11 for (int i = 0; i < 32; ++i) { 12 result |=bitsum[i] % 3 << i; 13 } 14 return result; 15 } 16 };
原文：http://www.cnblogs.com/skycore/p/3996533.html