LeetCode | 17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

两种方法,一种是递归(3ms),另一种看的大神解答,用迭代(0ms)

//递归
class Solution {
public:
        vector<char> tmp;
        vector<string> res;
        vector<string> Hash;

        void dfs(string digits,int index,int len)//处理第index位
        {
            if(index == len)
            {
                string t = "";
                for(int i=0;i<tmp.size();i++)
                {
                    t += tmp[i];
                }
                res.push_back(t);
                //tmp.erase(tmp.end()-1);
                return;
            }
            for(int i=0;i<Hash[digits[index]-'0'].length();i++)
            {
                tmp.push_back(Hash[digits[index]-'0'][i]);
                dfs(digits,index+1,len);
                tmp.erase(tmp.end()-1);//删除最后一个元素
            }
        }

        vector<string> letterCombinations(string digits)
        {
            int len = digits.length();
            if(len == 0)
            {
                return res;
            }
            Hash.push_back(""); Hash.push_back(""); Hash.push_back("abc"); Hash.push_back("def"); Hash.push_back("ghi");
            Hash.push_back("jkl"); Hash.push_back("mno"); Hash.push_back("pqrs"); Hash.push_back("tuv"); Hash.push_back("wxyz");
            dfs(digits,0,len);

            return res;
        }
};

//迭代
class Solution {
public:
        vector<string> letterCombinations(string digits) {
            vector<string> res;
            string charmap[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
            if(digits.size() == 0)
            {
                return res;
            }
            res.push_back("");
            for (int i = 0; i < digits.size(); i++)
            {
                vector<string> tmp;
                string chars = charmap[digits[i] - '0'];
                for (int c = 0; c < chars.size();c++)
                    for (int j = 0; j < res.size();j++)
                        tmp.push_back(res[j]+chars[c]);
                res = tmp;
            }
            return res;
        }
};