LeetCode - Refresh - Single Number

You can use a hash map to record the frequencys. Or you can use bit operation.

x ^ x = 0. So the only left is the one.

 

class Solution {
public:
    int singleNumber(int A[], int n) {
        int result = 0;
        for (int i = 0; i < n; i++) {
            x ^= A[i];
        }
        return result;
    }
};