C++实现LeetCode(17.电话号码的字母组合)

时间:2022-06-25 04:31:55

[LeetCode] 17. Letter Combinations of a Phone Number 电话号码的字母组合

Given a string containing digits from 2-9inclusive, 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"].

Note:

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

这道题让我们求电话号码的字母组合,即数字2到9中每个数字可以代表若干个字母,然后给一串数字,求出所有可能的组合。这里可以用递归 Recursion 来解,需要建立一个字典,用来保存每个数字所代表的字符串,然后还需要一个变量 level,记录当前生成的字符串的字符个数,实现套路和上述那些题十分类似。在递归函数中首先判断 level,如果跟 digits 中数字的个数相等了,将当前的组合加入结果 res 中,然后返回。我们通过 digits 中的数字到 dict 中取出字符串,然后遍历这个取出的字符串,将每个字符都加到当前的组合后面,并调用递归函数即可,参见代码如下:

解法一:

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class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        vector<string> res;
        vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        letterCombinationsDFS(digits, dict, 0, "", res);
        return res;
    }
    void letterCombinationsDFS(string& digits, vector<string>& dict, int level, string out, vector<string>& res) {
        if (level == digits.size()) {res.push_back(out); return;}
        string str = dict[digits[level] - '0'];
        for (int i = 0; i < str.size(); ++i) {
            letterCombinationsDFS(digits, dict, level + 1, out + str[i], res);
        }
    }
};

这道题也可以用迭代 Iterative 来解,在遍历 digits 中所有的数字时,先建立一个临时的字符串数组t,然后跟上面解法的操作一样,通过数字到 dict 中取出字符串 str,然后遍历取出字符串中的所有字符,再遍历当前结果 res 中的每一个字符串,将字符加到后面,并加入到临时字符串数组t中。取出的字符串 str 遍历完成后,将临时字符串数组赋值给结果 res,具体实现参见代码如下:

解法二:

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class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        vector<string> res{""};
        vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        for (int i = 0; i < digits.size(); ++i) {
            vector<string> t;
            string str = dict[digits[i] - '0'];
            for (int j = 0; j < str.size(); ++j) {
                for (string s : res) t.push_back(s + str[j]);
            }
            res = t;
        }
        return res;
    }
};

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原文链接:https://www.cnblogs.com/grandyang/p/4452220.html