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
int singleNumber(int* nums, int numsSize){
int i, sum = 0;
for (i = 0; i < numsSize; i++) {
sum ^= nums[i];
}
return sum;
}
不断异或，相同的元素异或出来是 0 ，0 与任何元素异或都是元素本身，由异或的交换律，整个数组元素异或完后就会只剩下那个单出来的元素。
