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.
用两个hash map,分别表示long to short (map1) 和 short to long (map2) ,并把要用到的大小写字母和数字用string表示出来待用
encode: 先check map1中是否有longURL, 如果存在直接返回对应的shortURL;如果没有,在string中随机选n个字符组合,并检查map2中是否存在该组合,如果存在再增加n个字符,直到该url组合是unique的。然后分别在map1, map2中都存入这对url pair
decode: 直接返回map2中对应long url即可
encode: time = O(n), space = O(n) n: # of urls
decode: time = O(1), space = O(1)
public class Codec { Map<String, String> long2short = new HashMap<>(); Map<String, String> short2long = new HashMap<>(); String ch = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz1234567890"; Random r = new Random(); // Encodes a URL to a shortened URL. public String encode(String longUrl) { if(long2short.containsKey(longUrl)) { return "http://tinyurl.com/" + long2short.get(longUrl); } StringBuilder sb = new StringBuilder(); while(short2long.containsKey(sb.toString())) { for(int i = 0; i < 6; i++) { sb.append(ch.charAt(r.nextInt(ch.length()))); } } long2short.put(longUrl, sb.toString()); short2long.put(sb.toString(), longUrl); return "http://tinyurl.com/" + sb.toString(); } // Decodes a shortened URL to its original URL. public String decode(String shortUrl) { return short2long.get(shortUrl.substring("http://tinyurl.com/".length())); } } // Your Codec object will be instantiated and called as such: // Codec codec = new Codec(); // codec.decode(codec.encode(url));