题目地址：https://leetcode.com/problems/keyboard-row/#/descriptionopen in new window
题目描述
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. 
Example :
Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]
题目大意
判断那些字符串能使用键盘中的其中一行就能全部拼出来。
解题方法
暴力解
暴力解决了。分别把三行弄在三个数组里，对于每个单词每个字母都去循环，数在三行中的个数分别多少。如果这个单词能在一张中打出来完，那么说明由某一行的个数为1，其他行都为0.
注意字符串数组的写法。
public class Solution {
public String[] findWords(String[] words) {
char []arr1 = new char[]{'q','w','e','r','t','y','u','i','o','p','Q','W','E','R','T','Y','U','I','O','P'};
char []arr2 = new char[]{'a','s','d','f','g','h','j','k','l','A','S','D','F','G','H','J','K','L'};
char []arr3 = new char[]{'z','x','c','v','b','n','m','Z','X','C','V','B','N','M'};
List<String> ans = new ArrayList<String>();
for(String word: words){
int count1 = 0, count2 = 0, count3 = 0;
for(int i =0; i < word.length(); i++){
for(int j =0; j < arr1.length; j++){
if(word.charAt(i) == arr1[j]){
count1++;
}
}
for(int j =0; j < arr2.length; j++){
if(word.charAt(i) == arr2[j]){
count2++;
}
}
for(int j =0; j < arr3.length; j++){
if(word.charAt(i) == arr3[j]){
count3++;
}
}
}
if((count1 != 0 && count2 == 0 && count3 == 0)
||(count1 == 0 && count2 != 0 && count3 == 0)
||(count1 == 0 && count2 == 0 && count3 != 0)){
ans.add(word);
}
}
String []answer = new String[ans.size()];
for(int i =0; i < ans.size(); i ++){
answer[i] = ans.get(i);
}
return answer;
}
}
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字典 + set
二刷，Python。
使用字典来保存字符串在哪一行，然后遍历每个字符串，看它所有的字符在哪几行，用set对行数去重，如果set的结果是1，说明可以使用一行就求解出来。
class Solution:
def findWords(self, words):
"""
:type words: List[str]
:rtype: List[str]
"""
rowdict = {}
for c in "qwertyuiopQWERTYUIOP":
rowdict[c] = 1
for c in "asdfghjklASDFGHJKL":
rowdict[c] = 2
for c in "zxcvbnmZXCVBNM":
rowdict[c] = 3
res = []
for word in words:
if len(set(rowdict[c] for c in word)) == 1:
res.append(word)
return res
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