问题描述: 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? 提示:bit manipulation(位操作) 参考:http…

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? 思路:位运算神题! 这一题我们将所有的数都看作是二进制表示形式. 因为所有的…

原题地址https://leetcode.com/problems/single-number-ii/ 题目描述Given an array of integers, every element appears three times except for one. Find that single one. 给出一个整数数组,除了某个元素外所有元素都出现三次.找出仅出现一次的数字. Note: 注意: Your algorithm should have a linear runtime co…

LeetCode 137. Single Number II(只出现一次的数字 II)…

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…

[LeetCode]137. Single Number II 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/single-number-ii/description/ 题目描述: Given an array of integers, every element appears three times except for one, which appears exactly once. Find that singl…

题目描述: 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? Single Number II Given…

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? 思路: 这个题是Single Number的进阶版,如果其他的数都出现两次而…

问题来源:Single Number II 问题描述:给定一个整数数组,除了一个整数出现一次之外,其余的每一个整数均出现三次,请找出这个出现一次的整数. 大家可能很熟悉另一个题目(Single Number):除了一个数出现一次之外,其余的均出现两次,找到出现一次的数.该问题很简单,大家肯定都知道解法:将所有的数异或,最后的结果即是出现一次的数.用到的知识是A^A=0,两个相同的数异或变为0,然后0^B=B,这样即可找到出现一次的数. 新问题有了变化,出现两次变成出现三次,整个问题的解法就不一样…

136. Single Number 除了一个数字,其他数字都出现了两遍. 用亦或解决,亦或的特点:1.相同的数结果为0,不同的数结果为1 2.与自己亦或为0,与0亦或为原来的数 class Solution { public: int singleNumber(vector<int>& nums) { if(nums.empty()) ; ; ;i < nums.size();i++) res ^= nums[i]; return res; } }; 137. Single N…