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算法的重要性，我就不多說了吧，想去大廠，就必須要經(jīng)過基礎(chǔ)知識和業(yè)務(wù)邏輯面試+算法面試。所以，為了提高大家的算法能力，這個公眾號后續(xù)每天帶大家做一道算法題，題目就從LeetCode上面選！

今天和大家聊的問題叫做電話號碼的字母組合?，我們先來看題面：

https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/

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.

題意

給定一個僅包含數(shù)字 2-9 的字符串，返回所有它能表示的字母組合。

給出數(shù)字到字母的映射如下（與電話按鍵相同）。注意 1 不對應(yīng)任何字母。

樣例

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

題解

這是個排列組合問題對吧？這個排列組合可以用樹的形式表示出來；

當(dāng)給定了輸入字符串，比如："23"，那么整棵樹就構(gòu)建完成了，如下：

問題轉(zhuǎn)化成了從根節(jié)點(diǎn)到空節(jié)點(diǎn)一共有多少條路徑；

class?Solution {
????public?List<String> letterCombinations(String?digits) {
????????// 數(shù)字與字母的對應(yīng)關(guān)系
????????String[] map = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
????????List<String> res = new?ArrayList<>();
????????if(digits.isEmpty())
????????????return?res;
????????// 圖中樹的根節(jié)點(diǎn)
????????res.add("");
????????
????????// 遍歷輸入的數(shù)字
????????for(char c : digits.toCharArray()){
????????????res = combine(map[c-'0'], res);
????????}
????????
????????return?res;
????}
????
????// 根據(jù)數(shù)字組合字母
????private?List<String> combine(String?digits, List<String> list){
????????List<String> res = new?ArrayList<>();

????????for(char c : digits.toCharArray()){
????????????for(String?s : list){
????????????????res.add(s+c);
????????????}
????????}
????????
????????return?res;
????}
}