LeetCode电话号码的字母组合（string（push_back/empty）/dfs）

20230130 顺利通过，是最简单的模板dfs题
20230201 顺利通过
20230204 顺利通过
20230218 dfs要return size为0要返回空字符串。
20230227 顺利通过
原题解

题目

约束

解法

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> combinations;
        if (digits.empty()) {
            return combinations;
        }
        unordered_map<char, string> phoneMap{
            {'2', "abc"},
            {'3', "def"},
            {'4', "ghi"},
            {'5', "jkl"},
            {'6', "mno"},
            {'7', "pqrs"},
            {'8', "tuv"},
            {'9', "wxyz"}
        };
        string combination;
        backtrack(combinations, phoneMap, digits, 0, combination);
        return combinations;
    }

    void backtrack(vector<string>& combinations, const unordered_map<char, string>& phoneMap, const string& digits, int index, string& combination) {
        if (index == digits.length()) {
            combinations.push_back(combination);
        } else {
            char digit = digits[index];
            const string& letters = phoneMap.at(digit);
            for (const char& letter: letters) {
                combination.push_back(letter);
                backtrack(combinations, phoneMap, digits, index + 1, combination);
                combination.pop_back();
            }
        }
    }
};

class Solution {
public:
    string tmp;
    vector<string> res;
    vector<string> board={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    void DFS(int pos,string digits){
        if(pos==digits.size()){
            res.push_back(tmp);
            return;
        }
        int num=digits[pos]-'0';
        for(int i=0;i<board[num].size();i++){
            tmp.push_back(board[num][i]);
            DFS(pos+1,digits);
            tmp.pop_back();
        }
    }
    vector<string> letterCombinations(string digits) {
        if(digits.size()==0) return res;
        DFS(0,digits);
        return res;
    }
};