14.leetCode500：keyboard row

题目：
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.

题意：给定一系列的字符串，找出那些字符串是由键盘的同一行上的字母组成的。

思路一：键盘上共有三行字母，每一行字母都映射到一个map中：map1、map2、map3。对每一个字符串，根据字符串的首字母确定在键盘的哪一行，随后逐一判断后面的字符是否属于同一行。

代码：

class Solution {    public static String[] findWords(String[] words) {        char[] c1 = {'q','w','e','r','t','y','u','i','o','p'};        char[] c2 = {'a','s','d','f','g','h','j','k','l'};        char[] c3 = {'z','x','c','v','b','n','m'};        Map<Character, Integer> map1 = new HashMap<Character, Integer>();        Map<Character, Integer> map2 = new HashMap<Character, Integer>();        Map<Character, Integer> map3 = new HashMap<Character, Integer>();        for(int i=0;i<c1.length;i++)            map1.put(c1[i],i);        for(int i=0;i<c2.length;i++)            map2.put(c2[i],i);        for(int i=0;i<c3.length;i++)            map3.put(c3[i],i);        List<String> ss = new ArrayList<String>();        for(String word:words){            String newWord = word.toLowerCase();            char[] cs = newWord.toCharArray();            char[] cs2 = word.toCharArray();            if(map1.containsKey(cs[0])){                String s = "";                for(int i=0;i<cs.length;i++){                    if(!map1.containsKey(cs[i]))                        break;                    else                        s = s+cs2[i];                    if(i == cs.length-1)                        ss.add(s);                }            }else if(map2.containsKey(cs[0])){                String s = "";                for(int i=0;i<cs.length;i++){                    if(!map2.containsKey(cs[i]))                        break;                    else                        s = s+cs2[i];                    if(i == cs.length-1)                        ss.add(s);                }            }else{                String s = "";                for(int i=0;i<cs.length;i++){                    if(!map3.containsKey(cs[i]))                        break;                    else                        s = s+cs2[i];                    if(i == cs.length-1)                        ss.add(s);                }            }        }          return (String[])ss.toArray(new String[ss.size()]);      }}

思路二：建立一个map，把键盘上的每一行字母和自己所在的行数映射起来。即qwertyuiop，每个字母都映射为0，asdfghjkl每个字母都映射成1，zxcvbnm每个字母都映射成2。对于需要判定的字符串，首先确定首字符所在的行数，并逐一判断后续字符是否与首字符属于同一行。若中间有不一致的，立马结束，否则，该字符串符合要求。

代码：

class Solution {    public static String[] findWords(String[] words) {        String[] keyboardWords = {"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"};        Map<Character,Integer> map = new HashMap<Character,Integer>();        for(int i=0;i<keyboardWords.length;i++)            for(char c:keyboardWords[i].toCharArray())                map.put(c,i);        List<String> finalWords = new ArrayList<String>();        for(String word: words){            char first = word.toUpperCase().charAt(0);            int index = map.get(first);            for(char c: word.toUpperCase().toCharArray()){                if(map.get(c)!=index){                   index = -1;                    break;                }            }            if(index!=-1)                finalWords.add(word);        }        return (String[])finalWords.toArray(new String[finalWords.size()]);    }}

思路一的执行时间为4ms，思路二的执行时间为6ms。