题目：

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

思路：

相关题目：

LeetCode OJ 之 Single Number (唯一的数字）

LeetCode OJ 之 Single Number II （唯一的数字 - 二）

参考：http://blog.csdn.net/u012243115/article/details/45286963。

代码：

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) 
    {
        vector<int> result;
        int len = nums.size();
        if(len < 2)
            return result;
            
        int tmp = 0;
        for(int i = 0 ; i < len ; i++)
            tmp ^= nums[i];
            
        int indexOfOne = 0;
        for(int i = 0 ; i < 32 ; i++)
        {
            if( (tmp & (1 << i) ) != 0) //注意：位操作符&的优先级很低，这里要加()。这里用1做移位操作比用tmp做移位操作更好，因为如果tmp是负数，右移的话前面会补1.（尽管这里不影响结果，但是最好还是不要对tmp做移位操作）
            {
                indexOfOne = i;
                break;
            }
        }
        int tmp2 = (1 << indexOfOne) ;
        int num1 = 0 , num2 = 0;
        for(int i = 0 ; i < len ; i++)
        {
            if(nums[i] & tmp2 )
                num1 ^= nums[i];
            else
                num2 ^= nums[i];
        }
        result.push_back(num1);
        result.push_back(num2);
        return result;
    }
};

LeetCode OJ 之 Single Number III （唯一的数字-三）

原文：http://blog.csdn.net/u012243115/article/details/47780705