[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;
    }
}

 

转载于:https://www.cnblogs.com/qiaomu/p/4678131.html