[LintCode] Letter Combinations of a Phone Number *的字母组合

[LintCode] Letter Combinations of a Phone Number *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.

[LintCode] Letter Combinations of a Phone Number *的字母组合

 Notice

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

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;
    }
};