17、Letter Combinations of a Phone Number

电话号码的字母组合

给定一个数字字符串,返回所有可能的字母组合,可以代表。

一个映射的数字字母(就像在电话里按钮)如下所示。

 注意事项

以上的答案是按照词典编撰顺序进行输出的，不过，在做本题时，你也可以任意选择你喜欢的输出顺序。

样例

给定 "23"

返回 ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]

我的代码：

class Solution {//递归形式，时间复杂度（3^n),空间复杂度（n)public:    vector<string> const T={"","","abc","def",                    "ghi","jkl","mno","pqrs","tuv","wxyz"};    vector<string> letterCombinations(string digits) {        vector<string> res;        if(digits.empty())return res;        dfs(res,"",0,digits);        return res;    }    void dfs(vector<string> &res,string s,int rank,string digits)    {        if(rank==digits.length())res.push_back(s);        else            for(auto c:T[digits[rank]-'0'])                dfs(res,s+c,rank+1,digits);        }};

经典代码：

// LeetCode, Letter Combinations of a Phone Number// 时间复杂度O(3^n)，空间复杂度O(1)class Solution {public:    const vector<string> keyboard { " ", "", "abc", "def", // '0','1','2',...            "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };    vector<string> letterCombinations (const string &digits) {        vector<string deep="9"> result(1, "");        for (auto d : digits) {            const size_t n = result.size();            const size_t m = keyboard[d - '0'].size();            result.resize(n * m);            for (size_t i = 0; i < m; ++i)                copy(result.begin(), result.begin() + n, result.begin() + n * i);            for (size_t i = 0; i < m; ++i) {                auto begin = result.begin();                for_each(begin + n * i, begin + n * (i+1), [&](string &s) {                    s += keyboard[d - '0'][i];                });            }        }        return result;    }};