——Letter Combinations of a Phone Number

17、Letter Combinations of a Phone Number

電話號碼的字母組合

給定一個數字字符串,返回所有可能的字母組合,可以代表。

一個映射的數字字母(就像在電話裏按鈕)如下所示。

 注意事項

以上的答案是按照詞典編撰順序進行輸出的，不過，在做本題時，你也可以任意選擇你喜歡的輸出順序。

樣例

給定 "23"

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

我的代碼：

class Solution {//遞歸形式，時間複雜度（3^n),空間複雜度（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);    
    }
};

經典代碼：

// LeetCode, Letter Combinations of a Phone Number
// 時間複雜度O(3^n)，空間複雜度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;
    }
};