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:
- The order of the result is not important. So in the above example,
[5, 3]
is also correct. - 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.
x &= -x 是用了树状数组中的lowbit。
1 class Solution { 2 public: 3 vector<int> singleNumber(vector<int>& nums) { 4 int x = 0; 5 for (auto n : nums) x ^= n; 6 int tmp = x, idx = 0; 7 x &= -x; 8 int a = 0, b = 0; 9 for (auto n : nums) { 10 if (n & x) a ^= n; 11 else b ^= n; 12 } 13 return {min(a, b), max(a, b)}; 14 } 15 };