Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
0和2个相同的数异或等于0
1 int singleNumber(int* nums, int numsSize) { 2 int flag = 0; 3 int i; 4 for(i = 0; i < numsSize; i++) 5 flag ^= nums[i]; 6 return flag; 7 }