Leetcode 17. Alphabet combination of phone numbers

Problem Description

Given numbers only a 2-9string, it returns all letter combinations indicated. The mapping from numbers to letters is given below (same as phone keys). Note 1 does not correspond to any letter.

Problem solving report

Typical DFS.

Set an index indexto determine whether digitsthe end has been searched .
In the dfs()interior, if you have to digitsend, this road will add the current generation of str to the result.
If it has not reached the digitsend, loop through the characters that can be added in the next step and increase the index indexby 1.

Implementation code

class Solution{
    public:
        map<char, string> M = {
            {'2', "abc"}, {'3', "def"}, {'4', "ghi"}, {'5', "jkl"}, 
            {'6', "mno"}, {'7', "pqrs"}, {'8', "tuv"}, {'9', "wxyz"}
        };
        vector<string> letterCombinations(string digits){
            if(digits.size()==0) return {};
            vector<string>ans;
            dfs(ans, "", 0, digits);
            return ans;
        }
        // 这个地方的 ans必须是值引用
        void dfs(vector<string>&ans, string curStr, int index, string digits){
            if(index==digits.size()) ans.push_back(curStr);
            else{
                for(int i=0;i<M[digits[index]].size();i++){
                    dfs(ans, curStr+M[digits[index]][i], index+1, digits);
                }
            }
        }
};

References

[1] Leetcode 17. The letter combination of the phone number

Leetcode 22. Bracket generation

Problem Description

The number n represents the logarithm of generating parentheses. Please design a function for generating all possible and effective combinations of parentheses.

Problem solving report

Similar to the previous question, omitted.

Implementation code

class Solution{
    public:
        vector<string> generateParenthesis(int n){
            if(n==0) return {};
            vector<string>ans;
            dfs(ans, "", n,0);
            return ans;
        }
        // 这个地方的 ans必须是值引用
        void dfs(vector<string>&ans, string target, int open, int close){
            if(open==0&&close==0) ans.push_back(target);
            else{
                if(open>0) dfs(ans, target+'(', open-1, close+1);
                if(close>0) dfs(ans, target+')', open, close-1);
            }
        }
};