Given an integer array 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. You can return the answer in any order.
Follow up: Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
Example 1:
Input: nums = [1,2,1,3,2,5] Output: [3,5] Explanation: [5, 3] is also a valid answer.
Example 2:
Input: nums = [-1,0] Output: [-1,0]
Example 3:
Input: nums = [0,1] Output: [1,0]
Constraints:
2 <= nums.length <= 3 * 104-231 <= nums[i] <= 231 - 1- Each integer in
numswill appear twice, only two integers will appear once.
class Solution:
def singleNumber(self, nums: List[int]) -> List[int]:
eor = 0
for num in nums:
eor ^= num
diff = eor & (~eor + 1)
a = 0
for num in nums:
if (num & diff) == 0:
a ^= num
b = eor ^ a
return [a, b]class Solution {
public int[] singleNumber(int[] nums) {
int eor = 0;
for (int num : nums) {
eor ^= num;
}
int diff = eor & (~eor + 1);
int a = 0;
for (int num : nums) {
if ((num & diff) == 0) {
a ^= num;
}
}
int b = eor ^ a;
return new int[]{a, b};
}
}