17. 电话号码的字母组合（回溯/DFS）

Posted 2022-12-01 Ston.V

1.Description

2.Example

 示例 1：

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
示例 2：

输入：digits = ""
输出：[]
示例 3：

输入：digits = "2"
输出：["a","b","c"]

3. Code

回溯法（DFS）:回溯都是一个模板，写几遍就晓得了。某处有多种选择时，for遍历；开始一轮后先检查边界条件收集结果，结束一轮后回退。

class Solution 
public:
    map<char,string> m= 
        '2',"abc",'3',"def",'4',"ghi",'5',"jkl",'6',"mno",'7',"pqrs",'8',"tuv",'9',"wxyz"
    ;
    int index =0;
    vector<string> res;
    string s = "";
    vector<string> letterCombinations(string digits) 
        int n = digits.length();
        if(n==0)
            return res;
        DFS(digits,index,s);
        return res;
    
    void DFS(string digits,int index,string &s)
        if(index == digits.length())
            res.push_back(s);
            return;
        else
            for(char c:m[digits[index]])
                s.push_back(c);
                DFS(digits,index+1,s);
                s.pop_back();
            
        

    
;

4.注意

1.string类型也有push_back和pop_back的操作