Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Example 1:
Input: nums = [2,2,1]
Output: 1
Example 2:
Input: nums = [4,1,2,1,2]
Output: 4
Example 3:
Input: nums = [1]
Output: 1
Constraints:
1 <= nums.length <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104- Each element in the array appears twice except for one element which appears only once.
Solution:
class Solution {
public int singleNumber(int[] nums) {
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < nums.length; i++){
if (!set.contains(nums[i])){
set.add(nums[i]);
}else{
set.remove(nums[i]);
}
}
for (int i : set){
return i;
}
return -1;
}
}
// TC: O(n)
// SC: O(n)
class Solution {
public int singleNumber(int[] nums) {
int result = 0;
for (int i = 0; i < nums.length; i++){
result = nums[i] ^ result;
}
return result;
}
}
// TC: O(n)
// SC: O(1)
// 0000000
// 1 ^ 1 xor = 0