[LeetCode] (medium) 260. Single Number III Medium

https://leetcode.com/problems/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.

Example:

Input:[1,2,1,3,2,5]
Output:[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?

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一路异或之后可以得到两个目标数纠缠在一起的状态，重点在于分离

一开始设想再以这个纠缠状态为起点一路异或一遍，当某个数使他为零时，就找到了第一个目标数，但是显然这个方法是错的

参考了 https://segmentfault.com/a/1190000004886431 

直观上理解就是一种将原数组的划分，分为两个子数组，每个数组中满足“有一个数出现一次，剩下的数出现两次”这一特性，就容易解了

注意取mask的时候类似于树状数组 -n = ~n+1 = ~(n-1)（证明思路：先证带符号数加法与无符号数相同，再证 n+~n+1 = 0，再证~n+1 = ~(n-1)，利用10...0后缀）

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        vector<int> result(2);
        
        int combine = 0;
        for(int cur : nums){
            combine ^= cur;
        }
        
        int mask = combine & (~(combine-1));
        
        int a = 0, b = 0;
        for(int cur : nums){
            if(cur & mask){
                a ^= cur;
            }else{
                b ^= cur;
            }
        }
        result[0] = a;
        result[1] = b;
        return result;
    }