[leetcode] 260. 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?

解法一：

依然是brute force的思路。

class Solution {public:    vector<int> singleNumber(vector<int>& nums) {        vector<int> res;        unordered_map<int,int> m;        for(int i=0; i<nums.size(); ++i)           m[nums[i]]++;                for(int i=0; i<nums.size();++i){            if(m[nums[i]]==1) res.push_back(nums[i]);        }        return res;            }};

解法二：

位操作是个重点。一开始的val其实是两个数字的亦或。val中为1的bit表明这两个数字在该bit上的值不同。这里要注意，val &= -val可以在val中只保留最后边为1的bit。比如11101000，就会得到00001000。于是，如果我们拿val &= -val的结果跟nums中的每一个数字做&运算，就可以把数字分成两组。每一次分别含有一个我们要求的数字。

class Solution {public:    vector<int> singleNumber(vector<int>& nums) {        vector<int> res(2,0);        int val = 0;        for(int i=0; i<nums.size();++i)            val^=nums[i];                val &= -val; //only keep the right 1-bit in val            for(int i=0; i<nums.size();++i){            if(nums[i]&val) res[0]^=nums[i];            else res[1]^=nums[i];        }        return res;                }};