Note: This is a companion problem to the System Design problem: Design TinyURL.
TinyURL is a URL shortening service where you enter a URL such as https://leetcode.com/problems/design-tinyurl and it returns a short URL such as http://tinyurl.com/4e9iAk.
Design the encode and decode methods for the TinyURL service. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.
C++:
class Solution {public: ???Solution() { ???????dict = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; ???????short2long.clear(); ???????long2short.clear(); ???????srand(time(NULL)); ???} ???// Encodes a URL to a shortened URL. ???string encode(string longUrl) { ???????if (long2short.count(longUrl)) { ???????????return "http://tinyurl.com/" + long2short[longUrl]; ???????} ???????int idx = 0; ???????string randStr; ???????for (int i = 0; i < 6; ++i) randStr.push_back(dict[rand() % 62]); ???????while (short2long.count(randStr)) { ???????????randStr[idx] = dict[rand() % 62]; ???????????idx = (idx + 1) % 5; ???????} ???????short2long[randStr] = longUrl; ???????long2short[longUrl] = randStr; ???????return "http://tinyurl.com/" + randStr; ???} ???// Decodes a shortened URL to its original URL. ???string decode(string shortUrl) { ???????string randStr = shortUrl.substr(shortUrl.find_last_of("/") + 1); ???????return short2long.count(randStr) ? short2long[randStr] : shortUrl; ???} ???private: ???unordered_map<string, string> short2long, long2short; ???string dict;};
[LeetCode] 535. Encode and Decode TinyURL 编码和解码短URL
原文地址：https://www.cnblogs.com/lightwindy/p/8481759.html