Find that single one.(linear runtime complexity0

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 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·

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