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 }