LeetCode 17 Letter Combinations of a Phone Number（电话号码的字母组合）

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翻译

给定一个数字字符串，返回所有这些数字可以表示的字母组合。一个数字到字母的映射（就像电话按钮）如下图所示。输入：数字字符串“23”输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]备注:尽管以上答案是无序的，如果你想的话你的答案可以是有序的。

原图

这里写图片描述

原文

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.

代码

看样子我还是用C#顺手点，一气呵成……主要是用递归，因为不知道digits的长度到底是多少，不可能写无数个foreach去判断。

public class Solution{    IList<string> list = new List<string>();    public Solution()    {        list.Insert(0, "abc");        list.Insert(1, "def");        list.Insert(2, "ghi");        list.Insert(3, "jkl");        list.Insert(4, "mno");        list.Insert(5, "pqrs");        list.Insert(6, "tuv");        list.Insert(7, "wxyz");    }    public IList<string> LetterCombinations(string digits)    {        IList<string> result = new List<string>();        if (digits.Length == 0)            return result;        if (digits.Length == 1)        {            foreach (var a in list.ElementAt(int.Parse(digits[0].ToString()) - 2))            {                result.Insert(0, a.ToString());            }        }        int count = 0;        IList<string> temp = LetterCombinations(digits.Substring(1, digits.Length - 1));        foreach (var a in list.ElementAt(int.Parse(digits[0].ToString()) - 2))        {            foreach (var rest in temp)            {                result.Insert(count++, a.ToString() + rest);            }        }        return result;    }}

以下是复制来的一段C++代码，坦白地说，我写不出来这样的C++代码……

class Solution {public:    vector<string> letterCombinations(string digits) {                vector<string> ans;        if(digits.size() == 0)            return ans;        int depth = digits.size();        string tmp(depth, 0);        dfs(tmp, 0, depth, ans, digits);        return ans;    }    void dfs(string &tmp, int curdep, int depth, vector<string> &ans, string &digits){        if(curdep >= depth){            ans.push_back(tmp);            return ;        }        for(int i = 0; i < dic[digits[curdep] - '0'].size(); ++ i){            tmp[curdep] = dic[digits[curdep] - '0'][i];            dfs(tmp, curdep + 1, depth, ans, digits);        }        return ;    }private:string dic[10] = {{""},{""},{"abc"},{"def"},{"ghi"},{"jkl"},{"mno"},{"pqrs"},{"tuv"},{"wxyz"}};};

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