LintCode-425: Letter Combinations of a Phone Number (DFS经典题)

电话号码的字母组合算法

原创已于 2022-11-10 17:15:29 修改 · 229 阅读

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于 2018-07-11 13:47:22 首次发布

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本文介绍了一种解决电话号码字母组合问题的经典DFS算法。通过递归的方式，结合电话按键对应的字母，生成所有可能的字母组合。此外，还讨论了使用BFS方法的可能性，并提出了在实时输出可能单词情况下的解决方案。

425 · Letter Combinations of a Phone Number
Algorithms
Medium
Description
Given a digit string excluded 0 and 1, 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.

1 2
ABC 3
DEF
4
GHI 5
JKL 6
MNO
7
PQRS 8
TUV 9
WXYZ

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

Example
Example 1:

Input: “23”
Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”]
Explanation:
‘2’ could be ‘a’, ‘b’ or ‘c’
‘3’ could be ‘d’, ‘e’ or ‘f’
Example 2:

Input: “5”
Output: [“j”, “k”, “l”]
Tags
Company
Amazon
Facebook
Dropbox
Uber
Google

经典DFS题。注意不要搞糊涂了用两层for循环。

注意这题也可以用BFS，下次补充。

要注意的是follow up。如果问在输入"23"的时候能否实时输出可能的单词，则要用Trie(Prefix tree)或Hashmap来预处理。用Hashmap的话，把所有的字母的前缀对应的单词放到Hashmap里面。但Hashmap的方法费空间，prefix的方法更好。

class Solution {
public:
    /**
     * @param digits: A digital string
     * @return: all posible letter combinations
     */
    vector<string> letterCombinations(string &digits) {
        map<int, string> hashmap = {{1,""}, {2,"abc"}, {3, "def"},
                                    {4, "ghi"}, {5, "jkl"}, {6, "mno"},
                                    {7, "pqrs"}, {8, "tuv"}, {9, "wxyz"}};
        vector<string> results;
        if (digits.empty()) return results;

        string sol;
        
        helper(digits, hashmap, 0, sol, results);
        return results;
    }

    void helper(string &digits, map<int, string> &hashmap, int index, string &sol, vector<string> &results) {
        if (index == digits.size()) {
            results.push_back(sol);
            return;
        }
        
        int digit = digits[index] - '0';
        
        for (int i = 0; i < hashmap[digit].size(); ++i) {
            sol.push_back(hashmap[digit][i]);
            helper(digits, hashmap, index + 1, sol, results);
            sol.pop_back();
        }
    }
};

二刷:

class Solution {
public:
    /**
     * @param digits: A digital string
     * @return: all posible letter combinations
     */
    vector<string> letterCombinations(string &digits) {
        if (digits.size() == 0) return {};
        string sol = "";
        vector<string> sols;
        helper(digits, 0, sol, sols); 
        return sols;
    }
private:
    map<char, vector<char>> dict = {{'2', {'a', 'b', 'c'}}, {'3', {'d', 'e', 'f'}}, 
                                   {'4', {'g', 'h', 'i'}}, {'5', {'j', 'k', 'l'}}, 
                                   {'6', {'m', 'n', 'o'}}, {'7', {'p', 'q', 'r', 's'}}, 
                                   {'8', {'t', 'u', 'v'}}, {'9', {'w', 'x', 'y', 'z'}}};
    
    void helper(string &digits, int pos, string sol, vector<string> &sols) {
        if (pos == digits.size()) {
            sols.push_back(sol);
            return;
        }
        for (int i = 0; i < dict[digits[pos]].size(); i++) {
            helper(digits, pos + 1, sol + dict[digits[pos]][i], sols);
        }
        return;    
    }     
};