17. 电话号码的字母组合

 

---

class Solution {
public:
//利用回溯，DFS方法 
    void help(int index,string &tmp,string &digits,map&record,vector&res)
    {
        if(index==digits.size())
        {
            res.push_back(tmp);
            return;
        }
        for(int i=0;i letterCombinations(string digits) {
        
        vector res;
        //返回为[]  不是[""]
        if(digits=="")return res;

        //vector>record={{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'}};
        maprecord={
        {'2', "abc"}, {'3', "def"}, {'4', "ghi"}, {'5', "jkl"}, {'6', "mno"},
        {'7', "pqrs"}, {'8', "tuv"}, {'9', "wxyz"}};
        string tmp;
        help(0,tmp,digits,record,res);

        return res;
    }
};

---

class Solution {
public:
    //1. 用map记录每个数字按键对应的所有字母
    map M = {
        {'2', "abc"}, {'3', "def"}, {'4', "ghi"}, {'5', "jkl"}, {'6', "mno"},
        {'7', "pqrs"}, {'8', "tuv"}, {'9', "wxyz"}
    };
    //2. 存储最终结果和临时结果的变量
    vector ans;
    string current;

    //3. DFS函数，index是生成临时结果字串的下标，
    //每一个digits[index]数字对应临时结果current[index]的一位字母
    void DFS(int index, string digits) {
        //出口
        if(index == digits.size()) {
            ans.push_back(current);
            return;
        }
        //递归调用
        //对于当前输入的第index号数字(digits[index])，
        //枚举其对应的所有字母(M[digits[index]][i])
        for(int i = 0; i < M[digits[index]].size(); i++) {
            current.push_back(M[digits[index]][i]);     //临时结果压入一个字母
            DFS(index + 1, digits);         //以在当前位置压入该字母这一“情况”为前提，构造此“分支”的后续结果
            current.pop_back();             //状态还原，例如临时结果从 "ab" -> "a"，下一次循环尝试"ac" 
        }
    }

    vector letterCombinations(string digits) {
        if(digits.size() == 0) {
            return ans;
        }
        DFS(0, digits);
        return ans;
    }

};