Find that single one.(linear runtime complexity0

 public class Solution {
     public int singleNumber(int[] nums) {

         int temp = 0;
         for (int i=0;i<nums.length;i++)
         {
             temp = temp^nums[i];
         }
         return temp;
     }
 }

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·