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136. Single Number

2024-08-18 00:17:00   218人阅读

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

 

java（8ms）：排序，然后跟相邻数字都不同的就是single one

 1 public class Solution {
 2     public int singleNumber(int[] nums) {
 3         Arrays.sort(nums);
 4         if (nums.length == 1)
 5             return nums[0] ;
 6         if (nums[0] != nums[1]){
 7             return nums[0] ;
 8         }else if (nums[nums.length-1] != nums[nums.length-2]){
 9             return nums[nums.length-1] ;
10         }
11         for (int i = 1 ; i < nums.length-1 ; i++){
12             if (nums[i] != nums[i-1] && nums[i] != nums[i+1]){
13                 return nums[i] ;
14             }
15             
16         }
17         return 0 ;
18     }
19 }

 

java（1ms）：相同的数异或后为0，将所有的数异或，最后剩下single one

1 public class Solution {
2     public int singleNumber(int[] nums) {
3         for (int i = 1 ; i < nums.length ; i++){
4             nums[0] ^= nums[i] ;
5         }
6         return nums[0] ;
7     }
8 }

 

136. Single Number


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