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?
方法一：
首先计算nums数组中所有数字的异或，记为xor
令lowbit = xor & -xor，lowbit的含义为xor从低位向高位，第一个非0位所对应的数字
例如假设xor = 6（二进制：0110），则-xor为（二进制：1010，-6的补码）
则lowbit = 2（二进制：0010）
1 class Solution { 2 public: 3 vector<int> singleNumber(vector<int>& nums) { 4 int size=nums.size(); 5 if(size==0||nums.empty()) 6 return vector<int>(); 7 int diff=0; 8 for(auto &ele:nums) 9 diff^=ele; 10 vector<int> res(2,0); 11 diff&=-diff;//从低位向高位，第一个非0位所对应的数字 12 for(auto &ele:nums) 13 if(diff&ele) 14 res[0]^=ele; 15 else 16 res[1]^=ele; 17 return res; 18 } 19 };
方法二：
1 class Solution { 2 public: 3 vector<int> singleNumber(vector<int>& nums) { 4 int size=nums.size(); 5 if(size==0||nums.empty()) 6 return vector<int>(); 7 int sum=0; 8 for(auto &ele:nums) 9 sum^=ele; 10 vector<int> res(2,0); 11 //从低位向高位，第一个非0位所对应的数字 12 int n=1; 13 while((sum&n)==0) 14 n=n<<1; 15 for(auto &ele:nums) 16 if(n&ele) 17 res[0]^=ele; 18 else 19 res[1]^=ele; 20 return res; 21 } 22 };