17. Letter Combinations of a Phone Number

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

用迭代的方法，每次从3个字符中选择一个然后传给下一次迭代

index是遍历的digits的索引

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> output;
        if(digits.empty())
            return output;
        string dict[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        combination(digits,dict,0,"",output);
        return output;
    }
    void combination(string digits,string dict[],int index,string input,vector<string> &output){
        if(index == digits.size()){
            output.push_back(input);
            return;
        }
        string str = dict[digits[index] - '0'];
        for(int i = 0;i < str.size();i++)
            combination(digits,dict,index+1,input + str[i],output);
    }
};

 自己写的另一种写法：

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