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.

题意：

判断一个单词的每一个字母是否都来自键盘上的同一行，是的话就放在最终结果的数组中

算法思路：

1.首先用3个字符串数组存放三行的字母

2.然后将三行的字母放到map中，对map设置不同的行设置不同的值

3.取每个单词的首字母进行判断，如果在map中取出的值不相等，flag=0，如果都满足在数组集合中添加这个单词

4.将数组list转化为数组

注意：在取每个单词的字母时，取出的是char类型的，直接get键值拿到的是NULL，所以在前面加上“”用来匹配字符串

代码实现

package easy;import java.util.ArrayList;import java.util.HashMap;public class KeyboardRow {public static String[] findWords(String[] words) {        String[] row1 = {"q", "w", "e", "r", "t", "y", "u", "i", "o", "p"};        String[] row2 = {"a", "s", "d", "f", "g", "h", "j", "k", "l"};        String[] row3 = {"z", "x", "c", "v", "b", "n", "m"};        HashMap hashMap = new HashMap();        ArrayList list = new ArrayList();        //添加第一行        for(int i=0; i<row1.length; i++){        hashMap.put(row1[i], 1);        }        //添加第二行        for(int i=0; i<row2.length; i++){        hashMap.put(row2[i], 2);        }        //添加第三行        for(int i=0; i<row3.length; i++){        hashMap.put(row3[i], 3);        }                for(int i=0; i<words.length; i++){//遍历每一个单词        boolean flag = true;        for(int j=0; j<words[i].length(); j++){        if(hashMap.get("" + words[i].toLowerCase().charAt(0)) != hashMap.get("" + words[i].toLowerCase().charAt(j))){             flag = false;        break;        }        }        if (flag){        list.add(words[i]);        }        }        String[] result = new String[list.size()];        return (String[])list.toArray(result);    }public static void main(String[] args){String[] words = {"Hello", "Alaska", "Dad", "Peace"};String[] result = findWords(words);for(int i=0; i<result.length; i++){System.out.println(result[i]);}}}