标签：backtrack code dia 长度 return str oid mon 字符串
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.
本题不是很了解如何讲解，直接给代码吧：
1 public class Solution { 2 public List<String> letterCombinations(String digits) { 3 LinkedList<String> res = new LinkedList<String>(); 4 if(digits.length()==0) return res; 5 String[] words = new String[]{"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; 6 res.add(""); 7 for(int i=0;i<digits.length();i++){ 8 int num = Character.getNumericValue(digits.charAt(i)); 9 while(res.peek().length()==i){ 10 String s = res.remove(); 11 for(char c:words[num].toCharArray()){ 12 res.add(s+c); 13 } 14 } 15 } 16 return res; 17 } 18 }
本题可以用比较客观的方法来解决，回溯法。因为这个是解决一个问题的所有解或者任意一个解，剪枝函数是要在给定的数字字符上选择。回溯的内容自然便是所有字符串的可能性，且使用的是dfs的方法。本题就是从空的字符串一点一点往上加，直到长度和digits的长度一样为止。记住，回溯法要有终止条件。代码如下：
1 public class Solution { 2 public List<String> letterCombinations(String digits) { 3 List<String> res = new LinkedList<String>(); 4 if(digits.length()==0) return res; 5 String[] words = new String[]{"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; 6 backtracking("",res,digits,words,0); 7 return res; 8 } 9 public void backtracking(String s,List<String> res,String digits,String[] words,int cur){ 10 if(cur==digits.length()){ 11 res.add(s); 12 }else{ 13 String word = words[digits.charAt(cur)-‘0‘]; 14 for(int i=0;i<word.length();i++){ 15 backtracking(s+word.charAt(i),res,digits,words,cur+1); 16 } 17 } 18 } 19 }
17. Letter Combinations of a Phone Number
标签：backtrack code dia 长度 return str oid mon 字符串
原文地址：http://www.cnblogs.com/codeskiller/p/6389169.html
