LeetCode 500 - Keyboard Row

https://leetcode.com/problems/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.

http://www.cnblogs.com/grandyang/p/6421749.html
首先我们把键盘的三行字符分别保存到三个set中，然后遍历每个单词中的每个字符，分别看当前字符是否在三个集合中，如果在，对应的标识变量变为1，我们统计三个标识变量之和就知道有几个集合参与其中了

    vector<string> findWords(vector<string>& words) {
        vector<string> res;
        unordered_set<char> row1{'q','w','e','r','t','y','u','i','o','p'};
        unordered_set<char> row2{'a','s','d','f','g','h','j','k','l'};
        unordered_set<char> row3{'z','x','c','v','b','n','m'};
        for (string word : words) {
            int one = 0, two = 0, three = 0;
            for (char c : word) {
                if (c < 'a') c += 32;
                if (row1.count(c)) one = 1;
                if (row2.count(c)) two = 1;
                if (row3.count(c)) three = 1;
                if (one + two + three > 1) break;
            }
            if (one + two + three == 1) res.push_back(word);
        }
        return res;
    }

判断输入单词的字母集合是否为键盘某一行字母集合的子集

https://discuss.leetcode.com/topic/77773/short-easy-java-with-explanation

    public String[] findWords(String[] words) {
        String[] strs = {"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"};
        Map<Character, Integer> map = new HashMap<>();
        for(int i = 0; i<strs.length; i++){
            for(char c: strs[i].toCharArray()){
                map.put(c, i);//put <char, rowIndex> pair into the map
            }
        }
        List<String> res = new LinkedList<>();
        for(String w: words){
            if(w.equals("")) continue;
            int index = map.get(w.toUpperCase().charAt(0));
            for(char c: w.toUpperCase().toCharArray()){
                if(map.get(c)!=index){
                    index = -1; //don't need a boolean flag. 
                    break;
                }
            }
            if(index!=-1) res.add(w);//if index != -1, this is a valid string
        }
        return res.toArray(new String[0]);
    }

http://bookshadow.com/weblog/2017/02/05/leetcode-keyboard-row/

def findWords(self, words): """ :type words: List[str] :rtype: List[str] """ rs = map(set, ['qwertyuiop','asdfghjkl','zxcvbnm']) ans = [] for word in words: wset = set(word.lower()) if any(wset <= rset for rset in rs): ans.append(word) return ans

https://discuss.leetcode.com/topic/77754/java-1-line-solution-via-regex-and-stream

public String[] findWords(String[] words) {
    return Stream.of(words).filter(s -> s.toLowerCase().matches("[qwertyuiop]*|[asdfghjkl]*|[zxcvbnm]*")).toArray(String[]::new);
}