电话号码的字母组合

长度最大为4，所以用暴搜就好了，先将数字对应的字符串存储下来。用回溯写就好了。

class Solution {
    string maps[10]={"","","abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    void dfs(int u,string&path,vector<string>&ans,string &digits)
    {
        if(u==digits.length())
        {
            ans.push_back(path);
            return;
        }
        for(char c:maps[digits[u]-'0'])
        {
            path[u]=c;
            dfs(u+1,path,ans,digits);
        }
    }
public:
    vector<string> letterCombinations(string digits) {
        int n=digits.length();
        if(n==0)return{};
        vector<string>ans;
        string path(n,0);
        dfs(0,path,ans,digits);
        return ans;

    }
};