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"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

本题速度差别主要体现在建立字典地方法，下边是两种方法：第一种beats 50%，第二种beats 8%。

public class Solution {    public List<String> letterCombinations(String digits) {        List<String> result = new ArrayList<String>();        String[] map = new String[] { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };        char[] tmp = new char[digits.length()];        if(digits.length() == 0)            return result;        rec(digits, 0, tmp, map, result);        return result;    }    public void rec(String digits, int index, char[] tmp, String[] map, List<String> result){        if(index == digits.length()){            result.add(new String(tmp));            return;        }        char tmpChar = digits.charAt(index);        for(int i = 0; i < map[tmpChar - '0'].length(); i++){            tmp[index] = map[tmpChar - '0'].charAt(i);            rec(digits, index + 1, tmp, map, result);        }    }}

public class Solution {    public List<String> letterCombinations(String digits) {    ArrayList<String> result=new ArrayList<>();if(digits.length()==0)return result;HashMap<Integer, List<Character>> hashMap=new HashMap<>();for(int i=2;i<10;i++){ArrayList<Character> tem=new ArrayList<>();if(i==7){tem.add('p');tem.add('q');tem.add('r');tem.add('s');}else if (i==8) {tem.add('t');tem.add('u');tem.add('v');}else if (i==9) {tem.add('w');tem.add('x');tem.add('y');tem.add('z');}else {for(int j=0;j<3;j++){tem.add((char)((i-2)*3+'a'+j));}}hashMap.put(i, tem);}doFor(hashMap, result, new StringBuilder(), digits, 0);        return result;    }    void doFor(Map<Integer, List<Character>> map,List<String> list,StringBuilder builder,String string,int i){for(char c:map.get(string.charAt(i)-'0')){builder.append(c);if(i==string.length()-1){list.add(builder.toString());}else {doFor(map, list, builder, string, i+1);}builder.setLength(builder.length()-1);}    }}