leetcode004-Letter Combinations of a Phone Number_alice is texting bob using her phone. the mapping -优快云博客

本文链接：https://blog.youkuaiyun.com/Alice_am/article/details/78175052

本文介绍了一种使用递归算法来解决电话号码字符串的所有可能字母组合问题的方法。通过将问题分解为更小的子问题，该算法能够有效地生成所有可能的组合。

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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.

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.

这道题感觉多层循环暴力求解是行不通的，因为数字串的长度未知。那么如果我们每次按顺序选定一个数字代表的字母，然后只需要求剩下子串所有的组合可能，按这样的思路考虑，我觉得递归可行，于是我就尝试了以下。

以下是AC的代码

class Solution {
public:
    vector<string> signs = {"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    vector<string> recur_substr(char c, string s) {
        vector<string> a;
        int t = c-'0'-2;
        if (s.size() == 0) {
            for (int i = 0; i < signs[t].size(); i++) {
                string temp;
                temp += signs[t][i];
                a.push_back(temp);
            }
        } else {
            vector<string> latter;
            if (s.size() == 1) {
                latter  = recur_substr(s[0], "");
            } else {
                latter  = recur_substr(s[0], s.substr(1));
            }
            for (int i = 0; i < signs[t].size(); i++) {
                for (int j = 0; j < latter.size(); j++) {
                    a.push_back(signs[t][i]+latter[j]);
                }
            }
        }
        return a;
    }

    vector<string> letterCombinations(string digits) {
        vector<string> ans;
        if (digits.empty()) {
            return ans;
        }
        if (digits.size()==1) {
            ans = recur_substr(digits[0], "");
        } else {
            ans = recur_substr(digits[0], digits.substr(1));
        }
        return ans;
    }
};