电话号码的字母组合

LeetCode每日一题，17. Letter Combinations of a Phone Number

先看题目描述

大意就是手机上的数字与字母对应，给出只含数字的字符串，让我们返回对应字母的排列组合

算法与思路

一开始的思路是模拟数字一个个的添加进来，先是添加 2 对应的，于是 res 中就有 [“a”, “b”, “c”]，然后将 3 添加进来，因为 3 对应 3 个字母，先将 res * 3 变为 [“a”, “b”, “c”, “a”, “b”, “c”, “a”, “b”, “c”]，然后分别添加 3 个 “d”，3 个 “e”，3 个 “f”，res 变为 [“ad”, “bd”, “cd”, “ae”, “be”, “ce”, “af”, “bf”, “cf”]，然后返回 res 即可，就用代码模拟这个过程

算法源码

import java.util.*;

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new LinkedList<>();
        if (digits.length() == 0) return res;
        res.add("");
        HashMap<Character, String> phone = new HashMap<>();
        phone.put('2', "abc");
        phone.put('3', "def");
        phone.put('4', "ghi");
        phone.put('5', "jkl");
        phone.put('6', "mno");
        phone.put('7', "pqrs");
        phone.put('8', "tuv");
        phone.put('9', "wxyz");
        for (char c: digits.toCharArray()) {
            int cur = 0;
            int coun = res.size();
            int len = phone.get(c).length();
            for (int i = 1; i < len; i++) {
                for (int j = 0; j < coun; j++) {
                    res.add(res.get(j));
                }
            }
            for (int i = 0; i < len; i++) {
                for (int j = 0; j < coun; j++) {
                    res.set(cur, res.get(cur) + phone.get(c).charAt(i));
                    cur++;
                }
            }
        }
        return res;
    }
}

但这个代码的运行效率不行，后来看了题解才发现可以用回溯，且频繁的字符串组装可以使用 StringBuilder 下面是使用回溯和 StringBuilder 的代码

import java.util.*;

class Solution {
    private List<String> res = new LinkedList<>();
    private String letterMap[] = {
            " ",    //0
            "",     //1
            "abc",  //2
            "def",  //3
            "ghi",  //4
            "jkl",  //5
            "mno",  //6
            "pqrs", //7
            "tuv",  //8
            "wxyz"  //9
    };

    public List<String> letterCombinations(String digits) {
        if (digits.length() == 0) return res;
        StringBuilder sb = new StringBuilder();
        combination(digits, 0, sb);
        return res;
    }

    private void combination(String digits, int index, StringBuilder sb) {
        if (index == digits.length()) {
            res.add(sb.toString());
            return;
        }
        Character c = digits.charAt(index);
        String letters = letterMap[c - '0'];
        for (int i = 0; i < letters.length(); i++) {
            combination(digits, index + 1, sb.append(letters.charAt(i)));
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}