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"].

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：

public class Solution {
    public List<String> letterCombinations(String digits) {
        LinkedList<String> res = new LinkedList<String>();
        if(digits.length()==0) return res;
        String[] words = new String[]{"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        res.add("");
        for(int i=0;i<digits.length();i++){
            int num = Character.getNumericValue(digits.charAt(i));
            while(res.peek().length()==i){
                String s = res.remove();
                for(char c:words[num].toCharArray()){
                    res.add(s+c);
                }
            }
        }
        return res;
    }
}

 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：

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new LinkedList<String>();
        if(digits.length()==0) return res;
        String[] words = new String[]{"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        backtracking("",res,digits,words,0);
        return res;
    }
    public void backtracking(String s,List<String> res,String digits,String[] words,int cur){
        if(cur==digits.length()){
            res.add(s);
        }else{
            String word = words[digits.charAt(cur)-'0'];
            for(int i=0;i<word.length();i++){
                backtracking(s+word.charAt(i),res,digits,words,cur+1);
            }
        }
    }
}