• [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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/qiaomu/p/4678131.html
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