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. Subscribe to see which companies asked this question.
1. 根据系统设计的Estimation of the amount of data we need to store for the next couple of years, 我们应需要6位Base62的char来encode
2. assume 避免地址爆炸，相同的longUrl得到相同的shortUrl, 这需要一个额外的hashMap longToShort
3. 这里因为我们想保证都是6位的shortURL，所以采用random generate的方法；其他方法还可以是编号等等
public class Codec {
HashMap<String, String> hashToUrl = new HashMap<String, String>();
HashMap<String, String> urlToHash = new HashMap<String, String>();
String tinyUrlBase = "http://tinyurl.com/";
String characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
Random random = new Random();
// Encodes a URL to a shortened URL.
public String encode(String longUrl) {
if (urlToHash.containsKey(longUrl))
return tinyUrlBase + urlToHash.get(longUrl);
StringBuilder hash = new StringBuilder();
do {
for (int i=0; i<6; i++) {
hash.append(characters.charAt(random.nextInt(characters.length())));
}
} while (hashToUrl.containsKey(hash.toString()));
hashToUrl.put(hash.toString(), longUrl);
urlToHash.put(longUrl, hash.toString());
return tinyUrlBase + hash.toString();
}
// Decodes a shortened URL to its original URL.
public String decode(String shortUrl) {
return hashToUrl.get(shortUrl.substring(tinyUrlBase.length()));
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(url));