1: intsingleNumber(int A[], int n) { 2: int left = A[0]; 3:for(int i =1;i< n;i++) 4: { 5: left = left ^ A[i]; 6: } 7: return left; 8: }
classSolution {public:intsingleNumber(vector<int>&nums) {if(nums.empty())return0;intret = nums[0];for(inti =1; i < nums.size(); i ++) ret^=nums[i];returnret; } }; 解法三:先排序,再遍历找出孤异元素 classSolution {public:intsingleNumber(vector<int>&nums) { sort(nums.begin(), ...
var singleNumber = function(nums) { return nums.reduce((res,a)=>res^a,0); }; 参考资料 LeetCode- Bit Manipulation LeetCode总结(1) —— 位运算:blog.csdn.net/xsloop/ar blog.csdn.net/zhning12L 欢迎加入码蜂社算法交流群:天天一道算法题 扫描下方二维码或搜素“码蜂社”公众号,不怕错过好文章...
代码: classSolution{public:intsingleNumber(vector<int>&nums){inta[35]={0};intn=nums.size();for(inti=0;i<n;i++){intbit;for(intj=0;j<32;j++){bit=1<<j;if((nums[i]&bit)!=0)a[j]++;}}intans=0;for(inti=0;i<32;i++)if(a[i]%3!=0)ans+=(1<<i);returnans;}}; 第...
Can you solve this real interview question? Single Number - Given a non-empty array of integers nums, every element appears twice except for one. Find that single one. You must implement a solution with a linear runtime complexity and use only constant
【leetcode】数组中找出只出现一次的数字(Single Number),题目是这样说的:Givenanarrayofintegers,everyelementappears twice exceptforone.Findthatsingleone.Note:Youralgorithmshouldhavealinearruntimecomplexity.Couldyouimplementitwit
LeetCode 136 Single Number(仅仅出现一次的数字) 翻译 给定一个整型数组,除了某个元素外其余元素均出现两次。 1. 找出这个仅仅出现一次的元素。 备注: 你的算法应该是一个线性时间复杂度。 你能够不用额外空间来实现它吗? 原文 Given an array of integers, every element appears twice except for one. Find ...
leetcode 算法解析(一):260. Single Number III 260.Single Number II 原题链接 本题其实算是比较简单,在 leetcode 上也只是 medium 级别,ac 率也很高,最好先自己尝试,本文只是单纯的记录一下自己整体的思路; 在阅读本文章之前,最好先解锁本题的简单模式136.Single Number,这对理解本题有较大的帮助;...
下面给出这个问题的Java代码(可以直接在leetcode上提交,下同): 解法一: classSolution{publicintsingleNumber(int[]nums){intres=0;for(intx:nums){res^=x;}returnres;}} 解法二: classSolution{publicintsingleNumber(int[]nums){int[]s=newint[32];//位计数器for(inti=0;i<nums.length;++i){for(int...
class Solution{public:vector<int>singleNumber(vector<int>&nums){intmask=0,a=0,b=0,div=1;for(auto&num:nums)mask^=num;while(!(div&mask))div<<=1;for(auto&num:nums){if(num&div)a^=num;elseb^=num;}returnvector<int>{a,b};}}; ...