500. Keyboard Row

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 1:

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

 

Note:

  1. You may use one character in the keyboard more than once.
  2. You may assume the input string will only contain letters of alphabet.

 

判断一个字符串的所有字母是否在键盘的同一行

C++(3ms):

 1 class Solution {
 2 public:
 3     vector<string> findWords(vector<string>& words) {
 4         unordered_set<char> row1  {'q','w','e','r','t','y','u','i','o','p'} ;
 5         unordered_set<char> row2  {'a','s','d','f','g','h','j','k','l'} ;
 6         unordered_set<char> row3  {'z','x','c','v','b','n','m'} ;
 7         vector<unordered_set<char>> row  {row1,row2,row3} ;
 8         vector<string> res ;
 9         for(string word : words){
10             int r = -1 ;
11             for(int i = 0 ; i < 3 ; i++){
12                 if (row[i].count(tolower(word[0])))
13                     r = i ;
14             }
15             if (r == -1)
16                 continue ;
17             bool flag = true ;
18             for(int j = 1 ; j < word.size() ; j++){
19                 if (row[r].count(tolower(word[j])) == 0){
20                     flag = false ;
21                     break ;
22                 }
23             }
24             if (flag)
25                 res.push_back(word) ;
26         }
27         return res ;
28     }
29 };

 

Java(4ms):

 1 class Solution {
 2     public String[] findWords(String[] words) {
 3         String[] strs = {"qwertyuiop","asdfghjkl","zxcvbnm"} ;
 4         Map<Character,Integer> map = new HashMap<>() ;
 5         List<String> res = new LinkedList<>() ;
 6         for (int i = 0 ; i < strs.length ; i++){
 7             for (char c : strs[i].toCharArray()){
 8                 map.put(c,i) ;
 9             }
10         }
11         for(String word : words){
12             if (word.equals("")) continue ;
13             int index = map.get(word.toLowerCase().charAt(0)) ;
14             for(char c : word.toLowerCase().toCharArray()){
15                 if (map.get(c) != index){
16                     index = -1 ;
17                     break ;
18                 }
19             }
20             if (index != -1)
21                 res.add(word) ;
22         }
23         return res.toArray(new String[res.size()]) ;
24     }
25 }

 

posted @ 2017-10-23 10:15  __Meng  阅读(177)  评论(0)  编辑  收藏  举报