136. Single Number【LeetCode】異或運算子，演算法，java

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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?

題目分析：數字每一位存在數組中，每個數字承兌出現，只有一個數字只出現一次，我們知道異或運算子相同為0不同為1.

//按位與運算&

System.out.println( 0 & 0 ); / / 0 System.out.println( 0 & 1 ); / / 0 System.out.println( 1 & 1 ); / / 1 System.out.println( "===========" ); / / 按位或運算子| System.out.println( 0 | 0 ); / / 0 System.out.println( 0 | 1 ); / / 1 System.out.println( 1 | 1 ); / / 1 System.out.println( "===========" ); / / 異或運算子^ System.out.println( 0 ^ 0 ); / / 0 System.out.println( 0 ^ 1 ); / / 1 System.out.println( 1 ^ 1 ); / / 0 System.out.println( "===========" );

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

 

136. Single Number【LeetCode】異或運算子，演算法，java