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?
class Solution {
public:
int singleNumber(int A[], int n) {
for(int i = 1; i < n; ++i){
A[0] ^= A[i];
}
return A[0];
}
};
异或在这里就完美演绎了，所有数字在计算机都是0和1，相同的都将消失，留下的就是那个Single Number，我很喜欢这道题。