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:

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.

class Solution {
    public String[] findWords(String[] word) {
        String[] words = new String[word.length];
        for(int i = 0; i < word.length; i++) words[i] = word[i];
        ArrayList<String> list = new ArrayList();
        ArrayList<String> res = new ArrayList();
        if(words.length == 0 || words == null) return new String[]{};
        list.add("QWERTYUIOP");
        list.add("ASDFGHJKL");
        list.add("ZXCVBNM");
        Set<Character> s1 = new HashSet();
        for(char c: list.get(0).toCharArray()) s1.add(c);
        
        Set<Character> s2 = new HashSet();
        for(char c: list.get(1).toCharArray()) s2.add(c);
        
        Set<Character> s3 = new HashSet();
        for(char c: list.get(2).toCharArray()) s3.add(c);
        
        for(int i = 0; i < words.length; i++){
            words[i] = words[i].toUpperCase();
            int count = 0;
            int real = 0;
            if(s1.contains(words[i].charAt(0))) {count = 1; real = words[i].length();}
            if(s2.contains(words[i].charAt(0))) {count = 2; real = words[i].length() * 2;}
            if(s3.contains(words[i].charAt(0))) {count = 3; real = words[i].length() * 3;}
            
            for(int j = 1; j < words[i].length(); j++){
                
                char c = words[i].charAt(j);
                if(s1.contains(c)) count += 1;
                if(s2.contains(c)) count += 2;
                if(s3.contains(c)) count += 3;
            }
            if(count == real) res.add(word[i]);
        }
        String[] resu = new String[res.size()];
        for(int i = 0; i < res.size(); i++) resu[i] = res.get(i);
        return resu;
    }
}

论如何把简单问题复杂化

class Solution {
    public String[] findWords(String[] word) {
        String[] words = new String[word.length];
        for(int i = 0; i < word.length; i++) words[i] = word[i];
        ArrayList<String> list = new ArrayList();
        ArrayList<String> res = new ArrayList();
        if(words.length == 0 || words == null) return new String[]{};
        list.add("QWERTYUIOPqwertyuiop");
        list.add("ASDFGHJKLasdfghjkl");
        list.add("ZXCVBNMzxcvbnm");
 
        
        for(int i = 0; i < words.length; i++){
            int count = 0;
            int real = 0;
            int le = words[i].length();
            char first = words[i].charAt(0);
            if(list.get(0).indexOf(first) >= 0) {count = 1; real = le;}
            if(list.get(1).indexOf(first) >= 0) {count = 2; real = le * 2;}
            if(list.get(2).indexOf(first) >= 0) {count = 3; real = le * 3;}
            
            for(int j = 1; j < words[i].length(); j++){
                
                char c = words[i].charAt(j);
                if(list.get(0).indexOf(c) >= 0) count += 1;
                if(list.get(1).indexOf(c) >= 0) count += 2;
                if(list.get(2).indexOf(c) >= 0) count += 3;
            }
            if(count == real) res.add(word[i]);
        }
        String[] resu = new String[res.size()];
        for(int i = 0; i < res.size(); i++) resu[i] = res.get(i);
        return resu;
    }
}

版本2，不用set，用indexOf判断