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.

想了下可能的重復編碼問題,直接用HashMap做了短URL和長URL的映射.將數字和字母混合編碼,按題目的6位數的編碼的話可以達到\(36^6\)種編碼,理論上性能是ok的.
代碼如下

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,""));
    }
}

寫完看了下排名.....貌似蠻低的.不過這題感覺是比較自由,應該不會太卡時間的,感覺就是考慮怎麽編碼好點.
前面的方案大致有:

1. 用計數器編碼
   那麽當前服務器存了多少url就曝露出來了，也許會有安全隱患。而且計數器編碼另一個缺點就是數字會不斷的增大
2. 直接返回參數(最快)
   ....這道題的OJ基本上是形同虛設，兩個函數分別直接返回參數字符串也能通過OJ，囧

Encode and Decode TinyURL