LeetCode 17 電話號碼的字母組合（難度：Medium）

題目大意：給定一個僅包含數字 2-9 的字符串，返回所有它能表示的字母組合。

給出數字到字母的映射如下（與電話按鍵相同）。注意 1 不對應任何字母。

 

 

Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

 

Example:

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

 

解決方案：回溯法

package pers.leetcode;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * Letter Combinations of a Phone Number（電話號碼的字母組合）
 * 難易程度：Medium
 *
 * @author jinghui.zhu
 * @date 2019/5/15 9:54
 */
public class LeetCode17 {

    public static void main(String[] args) {
        LeetCode17 leetCode17 = new LeetCode17();
        String phone = "23";
        System.out.println(leetCode17.letterCombinations(phone));

    }

    Map<String, String> phone = new HashMap<String, String>(){{
        put("2", "abc");
        put("3", "def");
        put("4", "ghi");
        put("5", "jkl");
        put("6", "mno");
        put("7", "pqrs");
        put("8", "tuv");
        put("9", "wxyz");
    }};

    List<String> output = new ArrayList<>();

    public void backtrack(String combination, String next_digits){
        if (next_digits.length() == 0){
            output.add(combination);
        }else {
            String digit = next_digits.substring(0, 1);
            String letters = phone.get(digit);
            for (int i = 0; i < letters.length(); i++){
                String letter = phone.get(digit).substring(i, i+1);
                backtrack(combination + letter, next_digits.substring(1));
            }
        }
    }

    public List<String> letterCombinations(String digits){
        if (digits.length() != 0){
            backtrack("", digits);
        }
        return output;
    }
}