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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()]);}这篇关于leetcode500 Keyboard Row Java的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
