Letter Combinations of a Phone Number 电话按键字符的组合@LeetCode

先搞一张电话数字和字符的对应表，再针对给定电话号码DFS回溯

package Level3;

import java.util.ArrayList;

/**
 * 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.
http://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Telephone-keypad2.svg/200px-Telephone-keypad2.svg.png


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.
 *
 */
public class S17 {

	public static void main(String[] args) {
		System.out.println(letterCombinations("23"));
	}

	public static ArrayList<String> letterCombinations(String digits) {
		String[] ss = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
		
		ArrayList<String> ret = new ArrayList<String>();
		rec(ret, digits.length(), ss, digits, new StringBuffer());
		return ret;
    }
	
	public static void rec(ArrayList<String> ret, int remain, String[] ss, String digits, StringBuffer sb){
		// 说明对digits遍历结束
		if(remain == 0){
			ret.add(sb.toString());
			return;
		}
		
		String s = ss[digits.charAt(0)-'0'];		// 得到digits[0]对应的string s
		// DFS
		for(int i=0; i<s.length(); i++){
			sb = sb.append(s.charAt(i));
			rec(ret, remain-1, ss, digits.substring(1), sb);
			sb.deleteCharAt(sb.length()-1);		// 恢复现场
		}
	}
}

public class Solution {
    
    public List<String> letterCombinations(String digits) {
        String[] map = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        List<String> list = new ArrayList<String>();
        rec(digits, map, list, new StringBuilder(), 0);
        return list;
    }
    
    public void rec(String digits, String[] map, List<String> list, StringBuilder sb, int pos) {
        if(pos == digits.length()) {
            list.add(sb.toString());
            return;
        }
        
        String s = map[digits.charAt(pos)-'0'];
        for(int i=0; i<s.length(); i++) {
            sb.append(s.charAt(i));
            rec(digits, map, list, sb, pos+1);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}