标签：
I title:
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：异或
class Solution { public: int singleNumber(vector<int>& nums) { int single = nums[0]; for (int i = 1 ;i < nums.size(); i++){ single ^= nums[i]; } return single; } };
II
title：
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?
思路：
这里我们需要重新思考，计算机是怎么存储数字的。考虑全部用二进制表示，如果我们把 第 ith 个位置上所有数字的和对3取余，那么只会有两个结果 0 或 1 (根据题意，3个0或3个1相加余数都为0). 因此取余的结果就是那个 “Single Number”.
一个直接的实现就是用大小为 32的数组来记录所有 位上的和。
class Solution { public: int singleNumber(vector<int>& nums) { vector<int> v(32,0); int result = 0; for (int i = 0; i < 32; i++){ for (int j = 0 ;j < nums.size(); j++){ if ((nums[j] >> i) & 1) v[i]++; } result |= ((v[i] % 3) << i); } return result; } };
这个算法是有改进的空间的，可以使用掩码变量：
ones 代表第ith 位只出现一次的掩码变量twos 代表第ith 位只出现两次次的掩码变量threes 代表第ith 位只出现三次的掩码变量假设在数组的开头连续出现3次5，则变化如下：
ones = 101 twos = 0 threes = 0 -------------- ones = 0 twos = 101 threes = 0 -------------- ones = 0 twos = 0 threes = 101 --------------
当第 ith 位出现3次时，我们就 ones 和 twos 的第 ith 位设置为0. 最终的答案就是 ones。
class Solution { public: int singleNumber(vector<int>& nums) { int one = 0, two = 0, three = 0; for (int i = 0; i < nums.size(); i++){ two |= (one & nums[i]); one ^= nums[i]; three = one & two; one &= ~three; two &= ~three; } return one; } };
LeetCode: Single Number I && II
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原文地址：http://www.cnblogs.com/yxzfscg/p/4524727.html
