[LeetCode]Single Number III

Single Number III

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

主要的问题是怎么把两个只有一个数的A和B分开。

如果全部异或的话得到的结果就是A^B。

A！=B，所以异或的结果中必然存在1，那么为1的那一位对应的A和B必然一个是1，一个是0，这个就可以将所有数字分为两类。

分别异或就能达到两个值了。

获取A,B最后不同的一位的方法是mask=AxorB & (~AxorB-1))。

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        vector<int> result;
        if(nums.size()<2) return result;
        int AxorB = 0;
        for(int i=0;i<nums.size();i++)
        {
            AxorB^=nums[i];
        }
        int mask = AxorB & (~(AxorB-1));
        int A=0,B=0;
        for(int i=0;i<nums.size();i++)
        {
            if(mask&nums[i])
            {
                A^=nums[i];
            }
            else
            {
                B^=nums[i];
            }
        }
        result.push_back(A);
        result.push_back(B);
        return result;
    }
};