leetcode 17. Letter Combinations of a Phone Number

/*leetcode 17. Letter Combinations of a Phone NumberGiven 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"].参考：http://www.cnblogs.com/ganganloveu/p/4175384.html枚举所有情况对于每一个输入数字，对于已有的排列中每一个字符串，分别加入该数字所代表的每一个字符。所有是三重for循环。举例：初始化排列{""}1、输入2，代表"abc"已有排列中只有字符串""，所以得到{"a","b","c"}2、输入3，代表"def"(1)对于排列中的首元素"a"，删除"a"，并分别加入'd','e','f'，得到{"b","c","ad","ae","af"}(2)对于排列中的首元素"b"，删除"b"，并分别加入'd','e','f'，得到{"c","ad","ae","af","bd","be","bf"}(2)对于排列中的首元素"c"，删除"c"，并分别加入'd','e','f'，得到{"ad","ae","af","bd","be","bf","cd","ce","cf"}参考 https://github.com/soulmachine/leetcode递归求解*/#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;class Solution {public:    vector<string> letterCombinations(string digits)     {        vector<string> ret;        if (digits == "")            return ret;        ret.push_back("");        vector<string> letters{ "", "", "abc", "def", "ghi", "jkl", "mno",            "pqrs", "tuv", "wxyz" };        for (size_t i = 0; i < digits.size(); ++i)        {            int size = ret.size();            for (size_t j = 0; j < size; ++j)            {                string cur = ret[0];        //取出第一个string                ret.erase(ret.begin());                //对于数字对应的字母来说，和cur相加，然后加入到数组中                for (size_t k = 0; k < letters[digits[i] - '0'].size(); ++k)                    ret.push_back(cur + letters[digits[i] - '0'][k]);            }        }        return ret;    }    //递归解法    //由于长度固定为digits.size(),那么每次到了这个长度就压入    const vector<string> letters{ "", "", "abc", "def", "ghi", "jkl", "mno",        "pqrs", "tuv", "wxyz" };    //digits为数字，cur为当前path的长度，path为找到的组合，reslut为结果数组    void dfs(const string& digits, size_t cur, string path, vector<string> result)    {        //当到了这个长度之后，说明path已经找到一个        if (cur == digits.size())        {            result.push_back(path);            return;        }        //如果没有，还要继续加入其他的字母。        for (auto c : letters[digits[cur] - '0'])            dfs(digits, cur + 1, path + c, result);    }    vector<string> letterCombinations1(string digits)    {        vector<string> result;        if (digits.empty())            return result;        dfs(digits, 0, "", result);        return result;    }};void TEST(){    Solution sol;    vector<string> ret;    vector<string> ret1;    ret = sol.letterCombinations("23");    ret1 = sol.letterCombinations1("23");    for (auto c : ret)        cout << c << " ";    cout << endl;    for (auto c : ret1)        cout << c << " ";    cout << endl;}int main(){    TEST();    return 0;}