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.
1 public class Solution { 2 public int[] singleNumber(int[] nums) { 3 // first pass to get xor result of all numbers 4 int temp = 0; 5 for (int num : nums) 6 temp ^= num; 7 8 // get the set bit 9 temp &= -temp; 10 11 int[] result = {0, 0}; 12 for(int num : nums) { 13 if ( (temp & num) == 0) { 14 result[0] ^= num; 15 } else { 16 result[1] ^= num; 17 } 18 } 19 return result; 20 } 21 }