[leetcode]Letter Combinations of a Phone Number

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.

 算法思路：

1.  很明显，这道题是组合问题combination，可以使用迭代来完成

2.  DFS。我的旧博客用的是combination的方法，不过当时心情不好，写的比较乱，这次我用的dfs，代码更简洁

public class Solution {
char[][] numberBoard = {{},{},{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},{'m','n','o'},{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList<String>();
        dfs(result,new StringBuilder(),digits);
        return result;
    }
    
    private void dfs(List<String> result,StringBuilder sb,String suffix){
        if(suffix.length() == 0){
            result.add(sb.toString());
            return;
        }
        int index = suffix.charAt(0) - '0';
        for(int i = 0; i < numberBoard[index].length; i++){
            dfs(result, sb.append(numberBoard[index][i]), suffix.substring(1));
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}

 第二遍：

对上一遍代码进行了一点点修复，如果digits中包含0或者1时，应该做出相应处理。

代码如下：

public class Solution {
    List<String> res = new ArrayList<String>();
    public List<String> letterCombinations(String digits) {
        dfs(new StringBuilder(),0,digits);
        return res;
    }
    private void dfs(StringBuilder sb,int k,String s){
        if(k >= s.length()){
            res.add(sb.toString());
            return;
        }
        char[][] board = {{},{},{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},{'m','n','o'},{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
        int n = s.charAt(k) - '0';
        if(n < 2) dfs(sb,k,s);//跳过，也可以抛出异常,总之需要对非法数字进行处理
        for(int i = 0; i < board[n].length; i++){
            sb.append(board[n][i]);
            dfs(sb,k + 1,s);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}