136. Single number "Leetcode" XOR operator, algorithm, Java

Given an array of integers, every element appears twice except for one. Find the single one.

Note:
Your algorithm should has a linear runtime complexity. Could you implement it without using extra memory?

Title Analysis: Each number of digits in the array, each digit acceptance appears, only one number appears once, we know that the XOR operator is the same as 0 different 1.

//按位与运算&

System.out.println( 0 & 0 ); / / 0 System.out.println( 0 & 1 ); / / 0 System.out.println( 1 & 1 ); / / 1 System.out.println( "===========" ); / / 按位或运算符| System.out.println( 0 | 0 ); / / 0 System.out.println( 0 | 1 ); / / 1 System.out.println( 1 | 1 ); / / 1 System.out.println( "===========" ); / / 异或运算符^ System.out.println( 0 ^ 0 ); / / 0 System.out.println( 0 ^ 1 ); / / 1 System.out.println( 1 ^ 1 ); / / 0 System.out.println( "===========" );

 Public class Solution {    publicint singlenumber (int[] nums) {        int result = 0;         int n =nums.length;          for (int i = 0; i<n; i++)        {            ^=nums[i];        }             return result;}    }

136. Single number "Leetcode" XOR operator, algorithm, Java