[leetcode] 535. Encode and Decode TinyURL

[leetcode] 535. Encode and Decode TinyURL

Description

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 ​​and it returns a short URL such as ​​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.

分析

题目的意思是：给定一个url，编码乘TinyURL，并且保证TinyURL能够解码回来。

这道题我也不怎么会，后面发现别人做得也是很简单的，encode阶段，如果该url已经编码了，则直接返回已经编码的url，如果没有，则生成随机数，然后判断随机数是否已经生成，如果生成了，则重新生成，然后存入url 的short long和long short的键值对

代码

class Solution {private: unordered_map short2long,long2short; string dict;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 "+ long2short[longUrl]; } string randStr; for(int i=0;i<6;i++){ randStr.push_back(dict[rand()%62]); } int idx=0; while(short2long.count(randStr)){ randStr[idx]=dict[rand()%62]; idx=(idx+1)%5; } short2long[randStr]=longUrl; long2short[longUrl]=randStr; return "+ long2short[longUrl]; } // 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; }};// Your Solution object will be instantiated and called as such:// Solution solution;// solution.decode(solution.encode(url));

参考文献

​​[LeetCode] Encode and Decode TinyURL 编码和解码精简URL地址​​

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