[Leetcode] Letter Combinations of a Phone Number

Letter Combinations of a Phone Number 题解

题目来源：https://leetcode.com/problems/letter-combinations-of-a-phone-number/description/

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Description

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.

Example

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.

Solution

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty())
            return vector<string>();
        vector<string> alphaMap = {
                " ",
               " ", "abc", "def",
               "ghi", "jkl", "mno",
               "pqrs", "tuv", "wxyz"
        };
        auto n = digits.size();
        vector<vector<string> > dp(n + 1);
        dp[0] = vector<string>{""};
        int i;
        for (i = 0; i < n; i++) {
            const string& alphas = alphaMap[digits[i] - '0'];
            for (auto c : alphas) {
                for (auto str : dp[i]) {
                    dp[i + 1].push_back(str + c);
                }
            }
        }
        return dp.back();
    }
};

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解题描述

这道题题意是给出一个数字串，然后根据数字串在九宫格键盘上的位置找出所有可能的字母组合。解法上用到的是类似哈希和DP的办法。

不过这道题还可以用回溯的办法来解决，可以减少空间消耗，时间复杂度是差不多的：

class Solution {
private:
    vector<string> alphaMap;
public:
    vector<string> letterCombinations(string digits) {
        vector<string> res;
        if (digits.empty())
            return res;
        alphaMap = {
                " ",
                " ", "abc", "def",
                "ghi", "jkl", "mno",
                "pqrs", "tuv", "wxyz"
        };
        string path;
        backTracking(res, path, digits, 0);
        return res;
    }

    void backTracking(vector<string>& res, string& path, const string& digits, int index) {
        if (index == digits.size()) {
            res.push_back(path);
        } else {
            const string& alphas = alphaMap[digits[index] - '0'];
            for (auto c : alphas) {
                path.push_back(c);
                backTracking(res, path, digits, index + 1);
                path.pop_back();
            }
        }
    }
};