leetcode500 Keyboard Row Java

Description

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

解法1

public class Solution {    public String[] findWords(String[] words) {        List<String> oneRowWords = new ArrayList<String>();        String[] keyboard = {"qwertyuiop","asdfghjkl","zxcvbnm"};        for(String word : words) {            String realWord = word;            word = word.toLowerCase();//每个字母变为小写            char[] strBit = word.toCharArray();            int count = 0;            for(char ch : strBit) {                if(keyboard[0].indexOf(strBit[0]) != -1) {//第一个字母在第一排                    if(keyboard[0].indexOf(ch) == -1) {//其他字母必须也在第一排 否则跳过                        break;                    }                }else if(keyboard[1].indexOf(strBit[0]) != -1) {//第一个字母在第二排                    if(keyboard[1].indexOf(ch) == -1) {                        break;                    }                }else if(keyboard[2].indexOf(strBit[0]) != -1) {//第一个字母在第三排                    if(keyboard[2].indexOf(ch) == -1) {                        break;                    }                }                count ++;            }            if(count == strBit.length) {                oneRowWords.add(realWord);            }        }        String[] oneRowWordsArray = new String[oneRowWords.size()];        for(int i=0; i<oneRowWords.size(); i++){            oneRowWordsArray[i] = oneRowWords.get(i);        }        return oneRowWordsArray;    }}

解法2

把键盘中的字母和其所在行数放到map。

public static String[] findWords2(String[] words) {        String[] Str = {"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"};        Map<Character,Integer> map = new HashMap<>();        for(int i=0; i<Str.length; i++) {            for(char c : Str[i].toCharArray()) {                map.put(c, i);            }        }        int index = 0;        List<String> res = new ArrayList<>();        for(String word : words) {            if (word.equals("")) continue;            index = map.get(word.toUpperCase().toCharArray()[0]);            for(char c : word.toUpperCase().toCharArray()) {                if(map.get(c) != index) {                    index = -1;//不用设置flag 直接把index设为-1即可                    break;                }            }            if(index != -1) res.add(word);        }        return res.toArray(new String[res.size()]);    }