[leetcode]Letter Combinations of a Phone Number

DFS，用C会更合适一些，用Java就要传很多的引用。在我看来，这个DFS就是递归。

import java.util.ArrayList;
import java.util.HashMap;

public class Solution {
    public ArrayList<String> letterCombinations(String digits) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<String> ans = new ArrayList<String>();
    	ArrayList<Character> chars = new ArrayList<Character>();
    	HashMap<Character, String> map = new HashMap<Character, String>();
    	init(map);
        dfs(map, digits, 0, chars, ans);
        if (ans.size() == 0) ans.add("");
        return ans;
    }
    
    private void dfs(HashMap<Character, String> map, String digits, int idx, ArrayList<Character> chars, ArrayList<String> arr)
    {
    	if (idx >= digits.length()) return;
    	char c = digits.charAt(idx);
    	String s = map.get(c);
    	for (int i = 0; i < s.length(); i++) {
    		
    		if (idx == digits.length() - 1) {
    			chars.add(s.charAt(i));
    			StringBuilder sb = new StringBuilder();
    			for (char cc : chars) {
    				sb.append(cc);
    			}
    			arr.add(sb.toString());
    			chars.remove(chars.size() - 1);
    		}
    		else
    		{
    			chars.add(s.charAt(i));
                dfs(map, digits, idx+1, chars, arr);
    			chars.remove(chars.size() - 1);
    		}
    	}
    }
    
    private void init(HashMap<Character, String> map)
    {
    	map.put('1', "");
    	map.put('2', "abc");
    	map.put('3', "def");
    	map.put('4', "ghi");
    	map.put('5', "jkl");
    	map.put('6', "mno");
    	map.put('7', "pqrs");
    	map.put('8', "tuv");
    	map.put('9', "wxyz");
    }
}

第二刷，这个其实类似于求全排列，适合DFS：

class Solution {
public:   
    vector<string> result;
    vector<string> letterCombinations(string digits) {
        string mapping[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        result.clear();
        string tmp;
        makeString(digits, mapping, tmp);
        return result;
    }
    
    void makeString(const string &digits, string mapping[], string &tmp)
    {
        int n = tmp.length();
        if (n == digits.length())
        {
            result.push_back(tmp);
            return;
        }
        int idx = digits[n] - '0' - 2;
        for (int i = 0; i < mapping[idx].length(); i++)
        {
            tmp.push_back(mapping[idx][i]);
            makeString(digits, mapping, tmp);
            tmp.pop_back();
        }
    }
};