【Simple】Lintcode 82: Single Number

Given 2*n + 1 numbers, every numbers occurs twice except one, find it.

Example

Given [1,2,2,1,3,4,3], return 4

Idea 1:

Violent solution, just sort first and then see the difference, although you can AC, but doing this the interviewer is going to beat people...

class Solution {
public:
    /**
     * @param A: An integer array
     * @return: An integer
     */
    int singleNumber(vector<int> &A)
    {
        // write your code here
        if(A.size() == 1) return A[0];
        
        sort(A.begin(),A.end());
        for(int i=0;i<A.size();i = i+2)
        {
            if(A[i] != A[i+1])
                return A[i];
        }
    }
};

Idea 2:

Through XOR, the result of the same number is 0, then the final result must be the number placed in the order.

3 characteristics of bitwise XOR:
(1) 0^0=0,0^1=1 0 XOR any number=any number
(2) 1^0=1,1^1=0 1 XOR any number - Negate any number

(3) XOR self of any number = set self to 0

The characteristic of XOR operation makes this method widely popularized to count the number of 1 digits of a number!

class Solution {
public:
    /**
     * @param A: An integer array
     * @return: An integer
     */
    int singleNumber(vector<int> &A)
    {
        // write your code here
        int years = 0;
        for(int i=0;i<A.size();i++)
            years ^= A[i];

        return ans;
    }
};