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.

这题好无聊，纯粹的重复性劳动，算法很简单就是DFS。

  1 public class Solution {
  2     ArrayList<String> result = null;
  3     public ArrayList<String> letterCombinations(String digits) {
  4         // Note: The Solution object is instantiated only once and is reused by each test case.
  5         result = new ArrayList<String>();
  6         if(digits == null) return result;
  7         if(digits.length() == 0){
  8             result.add("");
  9             return result;
 10         }
 11         getDigit(digits, 0, new StringBuffer());
 12         return result;
 13     }
 14     public void getDigit(String s,int pos,StringBuffer sb){
 15         if(pos == s.length()){
 16             result.add(sb.toString());
 17             return;
 18         }
 19         char c = s.charAt(pos);
 20         if(c == '2'){
 21             sb.append('a');
 22             getDigit(s,pos + 1, new StringBuffer(sb));
 23             sb.deleteCharAt(sb.length() - 1);
 24             
 25             sb.append('b');
 26             getDigit(s,pos + 1, new StringBuffer(sb));
 27             sb.deleteCharAt(sb.length() - 1);
 28             
 29             sb.append('c');
 30             getDigit(s,pos + 1, new StringBuffer(sb));
 31             sb.deleteCharAt(sb.length() - 1);
 32         }else if(c == '3'){
 33             sb.append('d');
 34             getDigit(s,pos + 1, new StringBuffer(sb));
 35             sb.deleteCharAt(sb.length() - 1);
 36             
 37             sb.append('e');
 38             getDigit(s,pos + 1, new StringBuffer(sb));
 39             sb.deleteCharAt(sb.length() - 1);
 40             
 41             sb.append('f');
 42             getDigit(s,pos + 1, new StringBuffer(sb));
 43             sb.deleteCharAt(sb.length() - 1);
 44         }else if(c == '4'){
 45             sb.append('g');
 46             getDigit(s,pos + 1, new StringBuffer(sb));
 47             sb.deleteCharAt(sb.length() - 1);
 48             
 49             sb.append('h');
 50             getDigit(s,pos + 1, new StringBuffer(sb));
 51             sb.deleteCharAt(sb.length() - 1);
 52             
 53             sb.append('i');
 54             getDigit(s,pos + 1, new StringBuffer(sb));
 55             sb.deleteCharAt(sb.length() - 1);
 56         }else if(c == '5'){
 57             sb.append('j');
 58             getDigit(s,pos + 1, new StringBuffer(sb));
 59             sb.deleteCharAt(sb.length() - 1);
 60             
 61             sb.append('k');
 62             getDigit(s,pos + 1, new StringBuffer(sb));
 63             sb.deleteCharAt(sb.length() - 1);
 64             
 65             sb.append('l');
 66             getDigit(s,pos + 1, new StringBuffer(sb));
 67             sb.deleteCharAt(sb.length() - 1);
 68         }else if(c == '6'){
 69             sb.append('m');
 70             getDigit(s,pos + 1, new StringBuffer(sb));
 71             sb.deleteCharAt(sb.length() - 1);
 72             
 73             sb.append('n');
 74             getDigit(s,pos + 1, new StringBuffer(sb));
 75             sb.deleteCharAt(sb.length() - 1);
 76             
 77             sb.append('o');
 78             getDigit(s,pos + 1, new StringBuffer(sb));
 79             sb.deleteCharAt(sb.length() - 1);
 80         }else if(c == '7'){
 81             sb.append('p');
 82             getDigit(s,pos + 1, new StringBuffer(sb));
 83             sb.deleteCharAt(sb.length() - 1);
 84             
 85             sb.append('q');
 86             getDigit(s,pos + 1, new StringBuffer(sb));
 87             sb.deleteCharAt(sb.length() - 1);
 88             
 89             sb.append('r');
 90             getDigit(s,pos + 1, new StringBuffer(sb));
 91             sb.deleteCharAt(sb.length() - 1);
 92             
 93             sb.append('s');
 94             getDigit(s,pos + 1, new StringBuffer(sb));
 95             sb.deleteCharAt(sb.length() - 1);
 96         }else if(c == '8'){
 97             sb.append('t');
 98             getDigit(s,pos + 1, new StringBuffer(sb));
 99             sb.deleteCharAt(sb.length() - 1);
100             
101             sb.append('u');
102             getDigit(s,pos + 1, new StringBuffer(sb));
103             sb.deleteCharAt(sb.length() - 1);
104             
105             sb.append('v');
106             getDigit(s,pos + 1, new StringBuffer(sb));
107             sb.deleteCharAt(sb.length() - 1);
108         }else if(c == '9'){
109             sb.append('w');
110             getDigit(s,pos + 1, new StringBuffer(sb));
111             sb.deleteCharAt(sb.length() - 1);
112             
113             sb.append('x');
114             getDigit(s,pos + 1, new StringBuffer(sb));
115             sb.deleteCharAt(sb.length() - 1);
116 
117             sb.append('y');
118             getDigit(s,pos + 1, new StringBuffer(sb));
119             sb.deleteCharAt(sb.length() - 1);
120             
121             sb.append('z');
122             getDigit(s,pos + 1, new StringBuffer(sb));
123             sb.deleteCharAt(sb.length() - 1);
124         }else if(c == '0'){
125             sb.append(' ');
126             getDigit(s,pos + 1, new StringBuffer(sb));
127             sb.deleteCharAt(sb.length() - 1);
128         }else{
129             return;
130         }
131     }
132 }

别人的解法：

 1 public class Solution {
 2     public ArrayList<String> letterCombinations(String digits) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         ArrayList<String> res = new ArrayList<String> ();
 6         res.add("");
 7         if(digits==null) return res;
 8         String[] table = digitToString();
 9         for(int i=0;i<digits.length();i++){
10             ArrayList<String> cur = new ArrayList<String>();
11             char c = digits.charAt(i);
12             if(c>='2' && c<='9'){
13                 for(String temp:res){
14                     for(int j=0;j<table[c-'2'].length();j++){
15                         cur.add(temp + table[c-'2'].charAt(j));
16                     }
17                 }
18             }
19             res= new ArrayList<String>(cur);
20         }
21         return res;    
22     }
23     public String[] digitToString(){
24       String[] res = new String[8];
25       char start = 'a';
26       
27       for(int i=0;i<8;i++){
28         int count = (i==5||i==7)?4:3;
29         StringBuilder temp = new StringBuilder("");
30         for(int j=0;j<count;j++){
31             temp.append((char)(start+j));
32         }
33         start=(char)(start+count);
34         res[i] = temp.toString(); 
35       }
36       return res;
37     }
38 }

第二遍：

 1 public class Solution {
 2     ArrayList<String> result = null;
 3     public ArrayList<String> letterCombinations(String digits) {
 4         // Note: The Solution object is instantiated only once and is reused by each test case.
 5          String[] c={   "",
 6                         "",
 7                         "abc",
 8                         "def",
 9                         "ghi",
10                         "jkl",
11                         "mno",
12                         "pqrs",
13                         "tuv",
14                         "wxyz"  };
15         result = new ArrayList<String>();
16         search( digits, 0, new StringBuffer(), c);
17         return result;
18     }
19     public void search(String s, int pos, StringBuffer sb,String[] c){
20         if(pos == s.length()){
21             result.add(sb.toString());
22             return;
23         }
24         int num = s.charAt(pos) - '0';
25         for(int i = 0; i < c[num].length(); i ++){
26             sb.append(c[num].charAt(i));
27             search(s, pos + 1, new StringBuffer(sb), c);
28             sb.deleteCharAt(sb.length() - 1);
29         }
30     }
31 }

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原文地址：https://www.cnblogs.com/reynold-lei/p/3371440.html