Leetccode 136 SingleNumber I 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? 1.自己想到了先排序,再遍历数…

Given an array of integers, every element appears twice except for one. Find that single one. 本题利用XOR的特性, X^0 = X, X^X = 0, 并且XOR满足交换律. class Solution(object): def singleNumber(self, nums): """ :type nums: List[int] :rtype: int ""…

题目 Given an array of integers, every element appears three times except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 分析 由上一题改动而来,LeetCode 136 Single Nu…

SingleNumber I: 题目链接:https://leetcode-cn.com/problems/single-number/ 题意: 给定一个非空整数数组,除了某个元素只出现一次以外,其余每个元素均出现两次.找出那个只出现了一次的元素. 说明: 你的算法应该具有线性时间复杂度. 你可以不使用额外空间来实现吗? 示例 1: 输入: [2,2,1] 输出: 1 示例 2: 输入: [4,1,2,1,2] 输出: 4 分析: 利用异或(xor)运算,两个相同的数异或为0 代码如下: cla…

https://leetcode.com/problems/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?…

136. Single Number 意思就是给你一堆数,每个数都出现了两次,只有一个数只出现了一次,找出这个数 位运算(和c艹一样) &:按位与 |:按位或 ^:异或(一样为0,不一样为1) 再说一下异或的性质,满足交换律和结合律 因此: 对于任意一个数n n ^ 0 = n n ^ n = 0 对于这道题来说,所有数依次异或剩下的就是那个数了 class Solution(object): def singleNumber(self, nums): ans = 0 for i in nums…

The question: 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? My analysis: Thi…

#include<stdio.h> #include<stdlib.h> int singleNumber(int* nums, int numsSize) { int count[32]={0}; int i,j,number=0; for(i=0;i<numsSize;i++) { for(j=0;j<32;j++) count[j]+=((nums[i]&(1<<j))!=0); } for(i=0;i<32;i++) { if(coun…

题目描述(面试常考题) 借助了异或的思想 class Solution { public: int singleNumber(vector<int>& nums) { ; ; i < nums.size() ; i++) ret = ret ^ nums[i]; return ret; /* 出现3次的情况借助了位运算的思想,相同的数 则 在int型的各32位都相等,所以32位中各位求和取余就可以得到结果 */ class Solution { public: int singl…

leetcode 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? 解题思路: 如果以线性复杂度和…