260.leetcode Single Number III(medium)[两个单数]

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.

1. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

数组中只有两个元素只出现了一次，其他元素都出现了两次，因此还是先做异或得到这两个元素异或的结果result，然后获取result最右边为1的位置。因为result为1的位置表示这两个单数在这一位上不同，因此可以按照在这一位上取值不同对数组再重新遍历异或得到r1和r2.

class Solution {public:    void FindFirstOne(int result,vector<int>& nums,int &r1,int& r2)    {         int pos = 0;        while((result&1) == 0)//寻找最右边是1的位置        {            pos++;            result = result>>1;        }        for(int i=0;i<nums.size();i++)        {            if((nums[i]>>pos)&1)            {                r1^=nums[i];            }else                r2^=nums[i];        }    }    vector<int> singleNumber(vector<int>& nums) {        if(nums.size()<=2) return nums;        int result = 0;        for(int i=0;i<nums.size();i++)            result ^=nums[i];        int r1=0,r2=0;        FindFirstOne(result,nums,r1,r2);        vector<int> last;        last.push_back(r1);        last.push_back(r2);        return last;    }};