【LeetCode】500. Keyboard Row

2019-11-10 22:53:04

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【鏈接】：keyboard-row 【描述】： Given a List of Words, return the words that can be typed using letters of alphabet on only one row’s of American keyboard like the image below.

這里寫(xiě)圖片描述 American keyboard

Example 1: Input: [“Hello”, “Alaska”, “Dad”, “Peace”] Output: [“Alaska”, “Dad”] Note: You may use one character in the keyboard more than once. You may assume the input string will only contain letters of alphabet. Subscribe to see which companies asked this question. 簡(jiǎn)單來(lái)說(shuō)就是找字符串，該字符串的每個(gè)字母都要在美式鍵盤(pán)的同一行即可。【思路】第一種方法三個(gè)set 集合存儲(chǔ)鍵盤(pán)布局，利用set的find功能查找是否滿足條件。第二種方法看了一下討論區(qū)，強(qiáng)大的python，居然有子集功能！【代碼】：

/***********************【LeetCode】500. Keyboard RowAuthor：herongweiTime:2017/2/8 13:10language：C++http://blog.csdn.net/u013050857***********************/#PRagma comment(linker,"/STACK:102400000,102400000")#include <bits/stdc++.h>#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1e5+10;const int maxm = 55;const LL MOD = 999999997;int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};inline LL read(){ int c=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();} return c*f;}/*Input: ["Hello", "Alaska", "Dad", "Peace"]//qwertyuiop//asdfghjkl//zxcvbnmOutput: ["Alaska", "Dad"]*/class Solution1 {public: vector<string> findWords(vector<string>& words) { int words_size=words.size(); vector<string> ret; unordered_set <char> row1 = { 'q','Q','w','W','e','E','r','R','t','T','y','Y','u','U','i','I','o','O','p','P' }; unordered_set <char> row2 = { 'a','A','s','S','d','D','f','F','g','G','h','H','j','J','k','K','l','L'}; unordered_set <char> row3 = { 'z','Z','x','X','c','C','v','V','b','B','n','N','m','M'}; for(auto &word:words){ bool f1=f2=f3=true; for(auto &ch:word){ if(f1){ auto it=row1.find(ch); if(it==row1.end()) f1=false; } if(f2){ auto it=row2.find(ch); if(it==row2.end()) f2=false; } if(f3){ auto it=row3.find(ch); if(it==row3.end()) f3=false; } } if(f1 || f2 || f3) ret.push_back(word); } return ret; }};class Solution2(object): def findWords(self, words): row1, row2, row3 = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm'); ret = []; for word in words: w = set(word.lower()); if w.issubset(row1) or w.issubset(row2) or w.issubset(row3): ret.append(word); return ret;

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