[LeetCode]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.

---

public class Solution {    public String[] findWords(String[] words) {        ArrayList<String> res = new ArrayList<String>();        String row0="QqWwEeRrTtYyUuIiOoPp";        String row1="AaSsDdFfGgHhJjKkLl";        String row2="ZzXxCcVvBbNnMm";        boolean[] whichr={false,false,false};        int count;                for(int i=0; i<words.length;i++){                       count=0;            for(int z=0;z< whichr.length;z++){                whichr[z]=false;            }                        for(int j=0;j<words[i].length();j++){                if(row0.indexOf(words[i].substring(j,j+1)) != -1) whichr[0]=true;                if(row1.indexOf(words[i].substring(j,j+1)) != -1) whichr[1]=true;                if(row2.indexOf(words[i].substring(j,j+1)) != -1) whichr[2]=true;            }            for(boolean ww : whichr){                if(ww) count++;            }                        if(count==1)                res.add(words[i]);        }                String[] ares = new String[res.size()];        for(int i=0;i<res.size();i++){            ares[i]=res.get(i);        }                for(String str : ares)System.out.println(str);                return ares;    }}

一些坑：

1、对数组，用arr.length

2、对String，用str.length()

3、下面这个赋值操作：

for(int z=0;z< whichr.length;z++){                whichr[z]=false;            }

一开始我是用这样：

for(boolean ww2 : whichr){            ww2=false;            }

这样是不对的，which里面的值没有被更新。

4、ArrayList的toArray

String[] array =new String[list.size()];list.toArray(array);

5、获取String里面的单独的char的方法

char[] strBit = word.toCharArray(); //变成char的数组char a1=word.charAt(j);  

另一种方法，用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()]);    }