标签：
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?
分析理解：
方法1：创建一个长度为int类型的位的个数的数组，也就是 sizeof( int ) * 8。用来一个数的各个位的1的个数，我们知道一个数出现3次，那么这个地方的位的1的个数肯定能整除3，这样就能找出只出现一次的那个数了。
// LeetCode, Single Number II
// 方法1，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
int singleNumber(int A[], int n) {
const int W = sizeof(int) * 8; // 一个整数的bit数，即整数字长
int count[W]; // count[i]表示在在i位出现的1的次数
fill_n(&count[0], W, 0);
for (int i = 0; i < n; i++) {
for (int j = 0; j < W; j++) {
count[j] += (A[i] >> j) & 1;
count[j] %= 3;
}
}
int result = 0;
for (int i = 0; i < W; i++) {
result += (count[i] << i);
}
return result;
}
};// LeetCode, Single Number II
// 方法2，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
int singleNumber(int A[], int n) {
int one = 0, two = 0, three = 0;
for (int i = 0; i < n; ++i) {
two |= (one & A[i]);
one ^= A[i];
three = ~(one & two);
one &= three;
two &= three;
}
return one;
}
};
LeetCode-Single Number II----位运算
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原文地址：http://blog.csdn.net/laojiu_/article/details/51356795
