136. Single Number - Java

136. Single Number

https://leetcode.com/problems/single-number/

Description

Given a non-empty 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?

Example 1:

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

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

Solutions

我的思路：
先排序，single number只会出现在第0、2、4等偶数（从零开始啊）位。

异或的方法：
we use bitwise XOR to solve this problem :

first , we have to know the bitwise XOR in java
若两个输入的电平相异，则输出为高电平（1）；若两个输入的电平相同，则输出为低电平（0）。
0 ^ N = N
N ^ N = 0
So… if N is the single number

N1 ^ N1 ^ N2 ^ N2 ^… Nx ^ Nx ^ N

= (N1^N1) ^ (N2^N2) ^… (Nx^Nx) ^ N

= 0 ^ 0 ^ …^ 0 ^ N

= N

Submissions

Arrays.sort():

class Solution {
    public int singleNumber(int[] nums) {
        Arrays.sort(nums);    
        for(int i=0;i<nums.length-1;i+=2)
        {
            if(nums[i]!=nums[i+1])            
                return nums[i];            
        }
        return nums[nums.length-1];
    }
}

XOR:

class Solution {
    public int singleNumber(int[] nums) {
        int result = 0;
        for (int num: nums) {
            result ^= num;
        }
        return result;
    }
}

Summary

Here’s the reason why it works. 1) XOR is commutative, a ^ b = b ^ a. 2) a number XOR by another number twice makes no change on original number, a ^ b ^ b = a
Therefore, if a number appears twice in the array, it makes no change to result. And if a number is unique, since 0 ^ unique = unique, so result = unique.