[LintCode] 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.

 Notice

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

Have you met this question in a real interview? 

Yes

Example

Given "23"

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

 

LeetCode上的原题，请参见我之前的博客Letter Combinations of a Phone Number。

 

解法一：

class Solution {
public:
    /**
     * @param digits A digital string
     * @return all posible letter combinations
     */
    vector<string> letterCombinations(string& digits) {
        if (digits.empty()) return {};
        vector<string> res;
        vector<string> v{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        helper(digits, v, 0, "", res);
        return res;
    }
    void helper(string& digits, vector<string>& v, int level, string out, vector<string>& res) {
        if (level == digits.size()) {
            res.push_back(out);
            return;
        }
        string t = v[digits[level] - '0'];
        for (int i = 0; i < t.size(); ++i) {
            out.push_back(t[i]);
            helper(digits, v, level + 1, out, res);
            out.pop_back();
        }
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param digits A digital string
     * @return all posible letter combinations
     */
    vector<string> letterCombinations(string& digits) {
        if (digits.empty()) return {};
        vector<string> res{""};
        vector<string> v{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        for (int i = 0; i < digits.size(); ++i) {
            string str = v[digits[i] - '0'];
            int n = res.size();
            for (int j = 0; j < n; ++j) {
                string t = res.front();
                res.erase(res.begin());
                for (int k = 0; k < str.size(); ++k) {
                    res.push_back(t + str[k]);
                }
            }
        }
        return res;
    }
};