17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string “23”
Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].
题意：九键输入法，有多少可能的字符串输出
代码解析：方法一：用迭代的思想，每一次把新的字符加入到已有的字符串上去再回溯到上一层
方法二耗时比较长，思想也是迭代，但是没有回溯

public class Solution3 {     public List letterCombinations(String digits) {         List result=new ArrayList();         StringBuffer sb=new StringBuffer();         int value=digits.length()-1;         if(digits.length()==0) {             return result;         }         letterCombinationsHelper(result,sb,0,value,digits);         return result;    }    public void letterCombinationsHelper(List<String> result,StringBuffer sb,int index,int max,String digits)    {        if(index>max)        {            result.add(new String(sb.toString()));            return;        }        char c=digits.charAt(index);        char start='a';        char end='c';        /* map character from phone number to alphabet*/        switch(c)        {            case '2':              start='a';              end='c';              break;            case '3':              start='d';              end='f';              break;              case '4':              start='g';              end='i';              break;            case '5':              start='j';              end='l';              break;            case '6':              start='m';              end='o';              break;            case '7':              start='p';              end='s';              break;            case '8':              start='t';              end='v';              break;            case '9':              start='w';              end= 'z';              break;         }        /*iterate over all the character mapping and back track*/        for(char val=start;val<=end;val++)        {            sb.append(val);            letterCombinationsHelper(result,sb,index+1,max,digits);            sb.deleteCharAt(sb.length()-1);        }    }}

方法二

public class Solution {    public List<String> letterCombinations(String digits) {        List<String> result = new ArrayList<String>();        int n = digits.length();        if(n < 1) return result;        result.add("");        for(int i = 0; i < n; i++)        {            char c = digits.charAt(i);            if(!dict.containsKey(c)) return new ArrayList<String>();            String s = dict.get(c);            List<String> temp = new ArrayList<String>();            for(String element : result)            {                for(char extra : s.toCharArray())                {                    temp.add(element + extra);                }            }            result = temp;        }        return result;    }   private static Map<Character,String> dict;    static    {        dict = new HashMap<Character, String>();        dict.put('2', "abc");        dict.put('3', "def");        dict.put('4', "ghi");        dict.put('5', "jkl");        dict.put('6', "mno");        dict.put('7', "pqrs");        dict.put('8', "tuv");        dict.put('9', "wxyz");    }}