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 x in nums:
eor ^= x
lowbit = eor & (-eor)
ans = [0, 0]
for x in nums:
if (x & lowbit) == 0:
ans[0] ^= x
ans[1] = eor ^ ans[0]
return ansclass Solution {
public int[] singleNumber(int[] nums) {
int eor = 0;
for (int x : nums) {
eor ^= x;
}
int lowbit = eor & (-eor);
int[] ans = new int[2];
for (int x : nums) {
if ((x & lowbit) == 0) {
ans[0] ^= x;
}
}
ans[1] = eor ^ ans[0];
return ans;
}
}/**
* @param {number[]} nums
* @return {number[]}
*/
var singleNumber = function(nums) {
let eor = 0;
for (const x of nums) {
eor ^= x;
}
const lowbit = eor & (-eor);
let ans = [0];
for (const x of nums) {
if ((x & lowbit) == 0) {
ans[0] ^= x;
}
}
ans.push(eor ^ ans[0]);
return ans;
};class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
long long eor = 0;
for (int x : nums) eor ^= x;
int lowbit = eor & (-eor);
vector<int> ans(2);
for (int x : nums)
if ((x & lowbit) == 0) ans[0] ^= x;
ans[1] = eor ^ ans[0];
return ans;
}
};func singleNumber(nums []int) []int {
eor := 0
for _, x := range nums {
eor ^= x
}
lowbit := eor & (-eor)
ans := make([]int, 2)
for _, x := range nums {
if (x & lowbit) == 0 {
ans[0] ^= x
}
}
ans[1] = eor ^ ans[0]
return ans
}