Find that single one.(linear runtime complexity0

 1 public class Solution {
 2     public int singleNumber(int[] nums) {
 3         
 4         int temp = 0;
 5         for (int i=0;i<nums.length;i++)
 6         {
 7             temp = temp^nums[i];
 8         }
 9         return temp;
10     }
11 }

 

 

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?

 

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相异为1；找到相等的部分；为1的部分是不同的·