[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 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.

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<string,string> 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 "http://tinyurl.com/" + 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 "http://tinyurl.com/" + 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

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

农民小飞侠

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值