Find the single one number

Problem description

Given a non-empty array of integers, every element appears twice except for one. Find that single one.Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Examples

Example 1:
Input: [2,2,1]
Output: 1

Example 2:
Input: [4,1,2,1,2]
Output: 4

Solution

　　异或概念： 　

&&	一假则假，短路
&	全判断
||	一真则真，短路
|	全判断
^(异或)	相异为真

　　剑指Offer的一道题，采用异或的方法。异或可以实现在不使用临时变量的情况下，交换两个值。

a = a ^ b
b = a ^ b // a ^ b ^ b = a ^ 0 = a
a = a ^ b // a ^ b ^ a = b ^ 0 = b

Code

 class Solution {
    public int singleNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
            result = result ^ nums[i];
        }
        return result;
    }

}