leetcode第17题——letter combination of a phone number

最新推荐文章于 2020-12-06 17:35:55 发布

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最新推荐文章于 2020-12-06 17:35:55 发布

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分类专栏： LeetCode刷题记录文章标签： leetcode 树 dfs

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本文介绍了一种使用深度优先搜索(DFS)算法解决电话号码字母组合问题的方法。通过构建数字到字母的映射，并利用递归搜索所有可能的组合，实现了快速求解。此算法在C++实现中表现出优异的时间和空间效率。

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问题描述
Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

Example:

Input: “23”
Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].
解法思路
基本想法是利用map建立一个图，然后利用dfs进行搜索，输出所有的路径

C++代码

 void recurseDfs(vector<string>& combos, string curCombo, string& digits, int digitsIdx) {
        static const char* phone[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        if (digits.length() == digitsIdx) {
            combos.push_back(curCombo);
            return;
        }
        for (const char *letters = phone[digits[digitsIdx]-'0']; *letters != '\0'; ++letters)
            recurseDfs(combos, curCombo+*letters, digits, digitsIdx+1);
    }
    
    vector<string> letterCombinations(string digits) {
        vector<string> combos;
        if (digits.length())
            recurseDfs(combos, "", digits, 0);
        return combos;
    }

结果分析
Runtime: 4 ms, faster than 100.00% of C++ online submissions for Letter Combinations of a Phone Number.
Memory Usage: 8.4 MB, less than 99.83% of C++ online submissions for Letter Combinations of a Phone Number.