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) {
        if(n<=0)
            cerr<<"Invalid input" <<endl;
        if(n==1)
            return A[0];
            
        int rst = A[0];
        for(int i = 1; i < n; ++i)
            rst = rst ^ A[i];
        return rst;
    }
};

TIPS：

1. 两两异或

2.C/C++的位运算

    &(按位与)、|(按位或)、^(按位异或)、~ (按位取反)。

 

转载于:https://www.cnblogs.com/nnoth/p/3744021.html