[LeetCode] Letter Combinations of a Phone Number 电话号码的字母组合

 

Given 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"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

 

这道题让我们求电话号码的字母组合，即数字2到9中每个数字可以代表若干个字母，然后给一串数字，求出所有可能的组合，相类似的题目有 Path Sum II 二叉树路径之和之二，Subsets II 子集合之二，Permutations 全排列，Permutations II 全排列之二，Combinations 组合项， Combination Sum 组合之和和 Combination Sum II 组合之和之二等等。我们用递归Recursion来解，我们需要建立一个字典，用来保存每个数字所代表的字符串，然后我们还需要一个变量level，记录当前生成的字符串的字符个数，实现套路和上述那些题十分类似，代码如下：

解法一

// Recursion

class Solution {

public:

    vector<string> letterCombinations(string digits) {

        vector<string> res;

        if (digits.empty()) return res;

        string dict[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

        letterCombinationsDFS(digits, dict, 0, "", res);

        return res;

    }

    void letterCombinationsDFS(string digits, string dict[], int level, string out, vector<string> &res) {

        if (level == digits.size()) res.push_back(out);

        else {

            string str = dict[digits[level] - '2'];

            for (int i = 0; i < str.size(); ++i) {

                out.push_back(str[i]);

                letterCombinationsDFS(digits, dict, level + 1, out, res);

                out.pop_back();

            }

        }

    }

};

 

这道题我们也可以用迭代Iterative来解，具体实现参见代码如下：

解法二

// Iterative

class Solution {

public:

    vector<string> letterCombinations(string digits) {

        vector<string> res;

        if (digits.empty()) return res;

        string dict[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

        res.push_back("");

        for (int i = 0; i < digits.size(); ++i) {

            int n = res.size();

            string str = dict[digits[i] - '2'];

            for (int j = 0; j < n; ++j) {

                string tmp = res.front();

                res.erase(res.begin());

                for (int k = 0; k < str.size(); ++k) {

                    res.push_back(tmp + str[k]);

                }

            }

        }

        return res;

    }

};