LeetCode每日一题：letter combination of a phone number

问题描述

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.

问题分析

枚举所有的情况，肯定是递归的。这题只要回溯即可。

代码实现

public ArrayList letterCombinations(String digits) {
        ArrayList result = new ArrayList<>();
        HashMap hashMap = new HashMap<>();
        hashMap.put('0', new char[]{' '});
        hashMap.put('2', new char[]{'a', 'b', 'c'});
        hashMap.put('3', new char[]{'d', 'e', 'f'});
        hashMap.put('4', new char[]{'g', 'h', 'i'});
        hashMap.put('5', new char[]{'j', 'k', 'l'});
        hashMap.put('6', new char[]{'m', 'n', 'o'});
        hashMap.put('7', new char[]{'p', 'q', 'r', 's'});
        hashMap.put('8', new char[]{'t', 'u', 'v'});
        hashMap.put('9', new char[]{'w', 'x', 'y', 'z'});
        getString(digits, 0, result, "", hashMap);
        return result;
    }

    private void getString(String digits, int position, ArrayList result,
                           String str, HashMap hashMap) {
        if (position < digits.length()) {
            if (hashMap.containsKey(digits.charAt(position))) {
                for (char c : hashMap.get(digits.charAt(position))) {
                    String newStr = str + c;
                    getString(digits, position + 1, result, newStr, hashMap);
                }
            }
        } else result.add(str);
    }