LeetCode Single Number

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?

思路分析：这题要求线性时间复杂度，需要用到位运算，X ^ X = 0，也就是任何数和自己取异或必然为0，于是这题我们可以把所有数组里的数取异或，最后剩下来的数就是那个只出现了一次的数。

AC Code：

public class Solution {
    public int singleNumber(int[] A) {
        int xorRes = 0;
        for(int i = 0; i < A.length; i++){
            xorRes = xorRes ^ A[i];
        }
        return xorRes;
    }
}