LeetCode 137. Single Number II--三次异或消除相同的数--C++,Python解法

题目地址：Single Number II - LeetCode

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

这是一道谷歌的面试题
这道题目意思很简单，就是找出没出现3次的数字。
最容易想到的想法是用字典，count。
Python解法如下：

from collections import Counter
class Solution:def singleNumber(self, nums: List[int]) -> int:s=Counter(nums)for i in s:if s[i]!=3:return i

然后会想可不可以用位运算，毕竟使用字典还是很费时的。
如何使用三次异或消去出现3次的数字。
官方解法如下：

class Solution:def singleNumber(self, nums: List[int]) -> int:seen_once = seen_twice = 0for num in nums:# first appearance: # add num to seen_once # don't add to seen_twice because of presence in seen_once# second appearance: # remove num from seen_once # add num to seen_twice# third appearance: # don't add to seen_once because of presence in seen_twice# remove num from seen_twiceseen_once = ~seen_twice & (seen_once ^ num)seen_twice = ~seen_once & (seen_twice ^ num)return seen_once

异或的解法难度极大，建议去看leetcode的讨论区：

C++解法如下：

class Solution {public:int singleNumber(vector<int>& nums) {int one = 0, two = 0, three = 0;for (int i = 0; i < nums.size(); ++i) {two |= one & nums[i];one ^= nums[i];three = one & two;one = one ^ three;two = two ^ three;}return one;}
};

LeetCode 137. Single Number II--三次异或消除相同的数--C++,Python解法相关推荐

【异或】LeetCode 137. Single Number II
LeetCode 137. Single Number II Solution1:不会做,抄的博客转载自:http://www.cnblogs.com/grandyang/p/4263927.htm ...
Leetcode 137. Single Number II JAVA语言
1 2 3 Given an array of integers, every element appears three times except for one, which appears ex ...
Leetcode #137 Single Number II
题目链接:https://leetcode.com/problems/single-number-ii/ 题意表明有一个数字出现了一次,其他数字出现了三次,那么实现一个模3的计数器即可. 1 clas ...
LeetCode 137.Single Number II 只出现一次的数字 II
题目描述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了三次.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: ...
137 Single Number II 数组中除了一个数外，其他的数都出现了三次，找出这个只出现一次的数...
给定一个整型数组,除了一个元素只出现一次外,其余每个元素都出现了三次.求出那个只出现一次的数. 注意: 你的算法应该具有线性的时间复杂度.你能否不使用额外的内存来实现? 详见:https://leet ...
LEETCODE 137.Singel Number ii
Given an array of integers, every element appears three times except for one. Find that single one. ...
[LeetCode]Single Number II
今天室友推荐了一道LeetCode上面的练习题Single Number II,题目大概的意思是有一个整形的数组,里面只有一个数是出现一次的,其他的数都出现了三次,然后让把这个数找出来. 第一眼看到这 ...
【？异或】LeetCode 260. Single Number III
LeetCode 260. Single Number III Solution1: 博客转载自:http://www.cnblogs.com/grandyang/p/4741122.html 这道题 ...
【异或】LeetCode 136. Single Number
LeetCode 136. Single Number Solution1:我的答案还好异或的性质没记错,还好,还好 class Solution { public:int singleNumber ...

LeetCode 137. Single Number II--三次异或消除相同的数--C++,Python解法

LeetCode 137. Single Number II--三次异或消除相同的数--C++,Python解法相关推荐

最新文章

热门文章