Encode and Decode TinyURL

Encode and Decode 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.

想了下可能的重复编码问题,直接用HashMap做了短URL和长URL的映射.将数字和字母混合编码,按题目的6位数的编码的话可以达到\(36^6\)种编码,理论上性能是ok的.

代码如下

public class Codec {

    HashMap<String, String> enIndex = new HashMap<String, String>();
    HashMap<String,String>deIndex = new HashMap<String, String>();
    String keyset = "0123456789abcdefghijklmnopqrstuvwyz";
    String tinyUrlBase = "http://tinyurl.com/";

    // Encodes a URL to a shortened URL.
    public String encode(String longUrl) {
        String key;
        if (enIndex.containsKey(longUrl)) return enIndex.get(longUrl);
        do{
            StringBuilder sb = new StringBuilder();
            for(int i =0 ; i<6;i++){
                sb.append(keyset.charAt((int) (Math.random()*keyset.length())));
            }
            key = sb.toString();
        }while(deIndex.containsKey(key));
        deIndex.put(key,longUrl);
        enIndex.put(longUrl, tinyUrlBase+key);
        return tinyUrlBase+key;
    }

    // Decodes a shortened URL to its original URL.
    public String decode(String shortUrl) {
        return deIndex.get(shortUrl.replace(tinyUrlBase,""));
    }
}

写完看了下排名.....貌似蛮低的.不过这题感觉是比较自由,应该不会太卡时间的,感觉就是考虑怎么编码好点.

前面的方案大致有:

1. 用计数器编码

   那么当前服务器存了多少url就曝露出来了，也许会有安全隐患。而且计数器编码另一个缺点就是数字会不断的增大

2. 直接返回参数(最快)

   ....这道题的OJ基本上是形同虚设，两个函数分别直接返回参数字符串也能通过OJ，囧