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 1: single-number Given an array of integers, every element appears twice except for one. Find that single one. solution: class solution: def singleNumber(self,A): A.sort() for i in range(1,len(A)): if i%2=1 && A[i]!=A[i-1]: return A[i-1] if __name__ == '...
分析: 数组中的数除了一个只出现了一次之外,其它都出现了两次, 要找出只出现了一次的数我们想到可以用XOR, 代码如下: publicclassSolution {publicintsingleNumber(int[] a) {intresult = a[0];for(inti=1;i<a.length;i++){ result^=a[i]; }returnresult; } }...
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...
classSolution{public:intsingleNumber(vector<int>&nums){intans=nums[0];intn=nums.size();for(inti=1;i<n;i++)ans=ans^nums[i];returnans;}}; 第二题题目:Given anon-emptyarray of integers, every element appearsthreetimes except for one, which appears exactly once. Find that single one.给...
【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
1. 2. 3. 4. 分析 请參照上一题:LeetCode 260 Single Number III(仅仅出现一次的数字3)(*) 另一道与之相应的题:LeetCode 137 Single Number II(仅仅出现一次的数字 II)(*) 代码 class Solution { public: unsigned int FindFirstBigIs1(int num) { ...
【C 語言的 LeetCode 30 天挑戰】第一天 (Single Number)是C 语言的 LeetCode 30 天挑战的第1集视频,该合集共计6集,视频收藏或关注UP主,及时了解更多相关视频内容。
于是利用交换律可以将数组假想成相同元素全部相邻,于是将所有元素依次做异或操作,相同元素异或为0,最终剩下的元素就为Single Number。时间复杂度O(n),空间复杂度O(1) publicintsingleNumber(int[] A){if(A ==null|| A.length ==0) {return-1;
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; ...