Day 23 - Keyboard Row

大家好，我是毛毛。ヾ(´∀ ˋ)ﾉ
废话不多说开始今天的解题Day~

500. Keyboard Row

Question

Given an array of strings words, return the words that can be typed using letters of the alphabet on only one row of American keyboard like the image below.

In the American keyboard:

the first row consists of the characters "qwertyuiop",
the second row consists of the characters "asdfghjkl", and
the third row consists of the characters "zxcvbnm".

Example

Example1

Input: words = ["Hello","Alaska","Dad","Peace"]
Output: ["Alaska","Dad"]

Example2

Input: words = ["omk"]
Output: []

Example3

Input: words = ["adsdf","sfd"]
Output: ["adsdf","sfd"]

Constraints

1 <= words.length <= 20
1 <= words[i].length <= 100
words[i] consists of English letters (both lowercase and uppercase).

解题

题目

首先先简单的翻译一下题目
给一组阵列的字串，要判断其中每个字使用到的字母是不是都在键盘上的同一列。

Think

作法大致上是这样

用一个dictionary存每个字母对应在键盘上的那一列，然後一个字一个字去判断只要其中有一个字跟前一个字的row不同就跳出判断这个字串的回圈，往下个字串找。
C再补上，先睡啦。

Code

Python

class Solution:
    def findWords(self, words: List[str]) -> List[str]:
        dict = {'Q':1, 'W':1, 'E':1, 'R':1, 'T':1, 'Y':1, 'U':1, 'I':1, 'O':1, 'P':1, 
               'A':2, 'S':2, 'D':2, 'F':2, 'G':2, 'H':2, 'J':2, 'K':2, 'L':2,
               'Z':3, 'X':3, 'C':3, 'V':3, 'B':3, 'N':3, 'M':3}
        ans = []
        flag = False
        
        for word in words:
            row = 0
            for index in range(len(word)):
                if len(word) == 1:
                    flag = True
                    
                if row != 0:
                    if row == dict[word[index].upper()]:
                        flag = True
                        continue
                    else:
                        flag = False
                        break
                else:
                    row = dict[word[index].upper()]
            
            if flag:
                ans.append(word)
            
        return ans