【leetcode】17 電話號碼組合

題目描述：

 https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/submissions/

 思路：

 代碼實現：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if(digits.empty()) return {};
        vector<string> res;//保存結果
        vector<string> dict{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};//字典0-9數字對應的字符串
 
        helper(digits, dict, 0,"", res);
        return res;
    }    

    void helper(string& digits, vector<string>& dict, int level, string out, vector<string>& res){
        //res存放最終結果，out存放臨時結果，level-生成的字符串大小，dict-字典
        if(level == digits.size()) {//如果level和digit大小相等，輸出out爲最終結果
            res.push_back(out);
            return ;
        }
        string str = dict[digits[level] - '0'];//找到digit中數字對應的字符串
        for(int i = 0; i < str.size(); i++){//遍歷字符串
            helper(digits, dict, level+1, out+str[i], res);
        }
    }
};

迭代：
class Solution {
public:
    vector<string> letterCombinations(string digits) {
       if (digits.empty()) return {};
        vector<string> res{""};
        vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
         for(int i = 0; i < digits.size(); i++){
            vector<string> t;
            string str = dict[digits[i] - '0'];
            for(int j = 0; j < str.size(); j++){
                for(string s:res)  t.push_back(s+str[j]);

            }
            res = t;
        }
        return res;
    }
};

參考：https://www.cnblogs.com/grandyang/p/4452220.html