Description

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

题意：给定一个数组，其中除了两个数出现一次外，其他都出现两次，要求在时间复杂度为O(n)空间复杂度为O(1)的条件下找出这两个数。

题解：按照异或的思想，最后可以得到这两个出现一次的数的异或值，那么我们只需对这个异或值lowbit操作，得到其最低位为1的位在哪，然后根据lowbit后得到的值，将数据分为两堆，那么可以知道的是这两个出现一次的数肯定分布在不同的两堆中（因为是对这这两个数的异或值lowbit操作，因此其为1的最低二进制位，这两个数肯定一个数0一个是1），那么问题转化为子问题，有一堆数，出了一个数出现一次外，其他数都出现两次。

 

class Solution {public:    //26ms    vector
     
       singleNumber(vector
      
       & nums) {        int diff = 0,one = 0,two = 0;        vector
       
        res;        for (unsigned int i = 0;i < nums.size();i++){            diff ^= nums[i];        }        diff &= (-diff);  //lowbit操作        for (unsigned int i = 0;i < nums.size();i++){            if (nums[i] & diff) one ^= nums[i];            else    two ^= nums[i];        }        res.push_back(one);        res.push_back(two);        return res;    }};