C++ 解法, 执行用时: 3ms, 内存消耗: 396KB, 提交时间: 2022-03-07

class Solution {
      string arr[10]={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param num string字符串 
     * @return string字符串vector
     */
   void _phone(const string& digits,int i,string combinstr,vector<string>&strv)
   {
       if(i==digits.size())
       {
        strv.push_back(combinstr);
        return;
       }
       string str=arr[digits[i]-'0'];
       for(int j=0;j<str.size();++j)
       {
          
        _phone(digits,i+1,combinstr+str[j],strv);


       }

   }
    vector<string> phoneNumber(string num) {
        // write code here
        string combinstr;
        vector<string> strv;
        if(num.empty())
        {
            return strv;
        }
       _phone(num,0,combinstr,strv);
        return strv;
        
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 412KB, 提交时间: 2022-07-29

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param num string字符串 
     * @return string字符串vector
     */
    
    string arr[10]={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    string tmp;
    vector<string> ans;
    vector<string> phoneNumber(string num) 
    {
        // write code here
        if(num.size()==0)
        {
            return ans;
        }
        backtrack(num,0);
        return ans;
        
    }
    
    void backtrack(string &num,int pos)
    {
        if(pos>=num.size())
        {
            ans.push_back(tmp);
            return ;
        }
        
        int n = num[pos]-'0';
        for(int i=0;i<arr[n].size();i++)
        {
            tmp+=arr[n][i];
            backtrack(num,  pos+1);
            tmp.pop_back();
        }
    }

};

C++ 解法, 执行用时: 3ms, 内存消耗: 412KB, 提交时间: 2022-06-29

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param num string字符串 
     * @return string字符串vector
     */
    const vector<string> lettermap={
        "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz",
    };
    vector<string> res;
    vector<string> phoneNumber(string num) {
        // write code here
        string s="";
        backtracing(num,s,0);
        return res;
    }
    void backtracing(string& num,string& s,int index){
         if(index==num.size()){
             res.push_back(s);
             return;
         }
        int digit=num[index]-'0';
        for(int i=0;i<lettermap[digit].size();++i){
            s.push_back(lettermap[digit][i]);
            backtracing(num,s,index+1);
            s.pop_back();
        }
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 416KB, 提交时间: 2022-07-11

class Solution {
    vector<vector<char>>map={
            {'A','B','C'},
            {'D','E','F'},
            {'G','H','I'},
            {'J','K','L'},
            {'M','N','O'},
            {'P','Q','R','S'},
            {'T','U','V'},
            {'W','X','Y','Z'}
        };
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param num string字符串 
     * @return string字符串vector
     */
    void dfs(vector<string>& ans,string t,string num,int i){
        if(i==num.length())ans.push_back(t);
        else{
            for(char c:map[num[i]-'2'])dfs(ans,t+c,num,i+1);
        }
    }
    vector<string> phoneNumber(string num) {
        // write code here
        vector<string>ans;
        dfs(ans,"",num,0);
        return ans;
    }
};

C++ 解法, 执行用时: 3ms, 内存消耗: 416KB, 提交时间: 2022-03-03

class Solution {
    
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param num string字符串 
     * @return string字符串vector
     */
        vector<string>ans;
    string tmp;
    const vector<string>dic={"ABC","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    vector<string> phoneNumber(string digits) {
        // write code here
                int len=digits.size();
        if(!len) return ans;
        backtrack(digits,0);
        reverse(ans.begin(),ans.end());
        return ans;
    }
        void backtrack(string &digits,int pos){
        if(pos>=digits.size()){
            ans.emplace_back(tmp);
            return;
        }
        int num=digits[pos]-'2';
        for(char c:dic[num]){
            tmp+=c;
            backtrack(digits,pos+1);
            tmp.pop_back();
        }
    }
};