[leedcode 136] Single Number

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?

public class Solution {
    public int singleNumber(int[] nums) {
        //一个数和他本身异或,为0
        int res=0;
        for(int i=0;i<nums.length;i++){
            res^=nums[i];
        }
        return res;
    }
}

 

posted @ 2015-07-26 17:01  ~每天进步一点点~  阅读(121)  评论(0编辑  收藏  举报