[Leetcode] letter combinations of a phone number 电话号码的字母组合

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.

 题意：给定数字串，输出对应字母的组合

思路：基本的DFS，这题和平常一般的DFS的区别，在于一般情况下，我们只需直接用下标进行深搜，而本题中，得先从数字串中取出一个数字，获得相应字符串以后，再在对应的字符串中获得一个字符，再在下一个数字对应的字符串中获得另一个字符，以这样的形式进行深搜整个过程类似于搜索二维数组，取数子相当于遍历行，取字母相当于遍历列。代码如下：

class Solution {
public:
    vector<string> letterCombinations(string digits) 
    {
        vector<string> res;
        if(digits.empty())  return {""};    //当digits为空的时候，应该返回""
        string dict[]={" "," ","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; 
        string midstr;
        letterDFS(digits,res,dict,midstr,0);
        return res;
    }

    void letterDFS(const string &digits,vector<string> &res,string dict[],string midstr, int start)
    {
        if(start==digits.size())    
            res.push_back(midstr);
        else
        {
            string str=dict[digits[start]-'0'];
            for(int i=start;i<str.size();++i)
            {
                midstr.push_back(str[i]);
                letterDFS(digits,res,dict,midstr,start+1);
                midstr.pop_back();
            }
        }
        

    }
};

方法二：这种方法的思路和subset的迭代解法很像。这里因为有一个取数字串中数字的情况存在，所以，多一个for循环，其余的基本一样。如：“23” ，res中先存入“a”,"b","c"，然后和“3”对应的字符相结合，后来存入"ad","ae","af",    "bd","be","bf",   "cd","ce","cf"参考了Cindy_niu的博客

class Solution {
public:
    vector<string> letterCombinations(string digits) 
    {
        vector<string> res(1,"");
        string dict[]={" "," ","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};

        for(int i=0;i<digits.size();++i)
        {
            vector<string> temp;
            for(int j=0;j<res.size();++j)
            {
                for(int k=0;k<dict[digits[i]-'0'].size();++k)
                {
                    temp.push_back(res[j]+dict[digits[i]-'0'][k]);
                }
            }
            res=temp;
        }     
        return res;
    }
};