leetcode 题解 || Letter Combinations of a Phone Number 问题

Posted 2020-09-14 cxchanpin

problem：

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.

 thinking：

（1）满足要求的数字为2~9。

（2）用一个array保存数字相应的字符，再用dfs枚举全部解

code：

class Solution {
private:
    map<char, vector<char> > dict;
    vector<string> ret;
public:
    void createDict()
    {
        dict.clear();
        dict[‘2‘].push_back(‘a‘);
        dict[‘2‘].push_back(‘b‘);
        dict[‘2‘].push_back(‘c‘);
        dict[‘3‘].push_back(‘d‘);
        dict[‘3‘].push_back(‘e‘);
        dict[‘3‘].push_back(‘f‘);
        dict[‘4‘].push_back(‘g‘);
        dict[‘4‘].push_back(‘h‘);
        dict[‘4‘].push_back(‘i‘);
        dict[‘5‘].push_back(‘j‘);
        dict[‘5‘].push_back(‘k‘);
        dict[‘5‘].push_back(‘l‘);
        dict[‘6‘].push_back(‘m‘);
        dict[‘6‘].push_back(‘n‘);
        dict[‘6‘].push_back(‘o‘);
        dict[‘7‘].push_back(‘p‘);
        dict[‘7‘].push_back(‘q‘);
        dict[‘7‘].push_back(‘r‘);
        dict[‘7‘].push_back(‘s‘);
        dict[‘8‘].push_back(‘t‘);
        dict[‘8‘].push_back(‘u‘);
        dict[‘8‘].push_back(‘v‘);
        dict[‘9‘].push_back(‘w‘);
        dict[‘9‘].push_back(‘x‘);
        dict[‘9‘].push_back(‘y‘);
        dict[‘9‘].push_back(‘z‘);
    }
    
    void dfs(int dep, int maxDep, string &s, string ans)
    {
        if (dep == maxDep)
        {
            ret.push_back(ans);
            return;
        }
        
        for(int i = 0; i < dict[s[dep]].size(); i++)
            dfs(dep + 1, maxDep, s, ans + dict[s[dep]][i]);
    }
    
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        if(digits.size()==0)
            return ret;
        createDict();
        dfs(0, digits.size(), digits, "");
        return ret;
    }
};