[Leetcode] Letter Combinations of a Phone Number

电话号码字母组合

最新推荐文章于 2020-08-10 10:20:18 发布

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最新推荐文章于 2020-08-10 10:20:18 发布

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原文链接：http://www.cnblogs.com/yanhewu/p/8397757.html

本文针对LeetCode中的Letter Combinations of a Phone Number问题提供了解决方案。介绍了如何利用类似哈希和动态规划的方法来生成所有可能的字母组合，并提供了一种使用回溯方法的替代方案。

Letter Combinations of a Phone Number 题解

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

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();
    }
};

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();
            }
        }
    }
};

转载于:https://www.cnblogs.com/yanhewu/p/8397757.html