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.
这里长网址变成短网址采用的方法是，定义一个常量字符串0-9a-zA-Z，每次将长网址作为key存入Map时，在将对应的短网址（在那个字符串常量中随机挑去6个作为短网址）作为对应的value存入map中，在将两者对换，存入另一个map中，用于将短网址变成长网址。存入过程中会判断长网址是否出现过，以及随机生成的短网址是否出现过，在做相应处理。
public class Codec { ???private static final String dict = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIKKLMNOPQRSTUVWXYZ"; ???Map<String,String> longToShort = new HashMap<>(); ???Map<String,String> shortToLong = new HashMap<>(); ???????// Encodes a URL to a shortened URL. ???public String encode(String longUrl) { ???????if(longToShort.containsKey(longUrl)){ ???????????return longToShort.get(longUrl); ???????} ???????String encoded = ?encodeString(); ???????while(shortToLong.containsKey(encoded)){ ???????????encoded = encodeString(); ???????} ???????longToShort.put(longUrl, encoded); ???????shortToLong.put(encoded, longUrl); ???????return encoded; ???} ???????public String encodeString(){ ???????String res = ""; ???????for(int i = 0; i<6; i++){ ???????????String str = String.valueOf(dict.charAt((int)(Math.random()*62))); ???????????res = res + str; ???????} ???????return res; ???} ???// Decodes a shortened URL to its original URL. ???public String decode(String shortUrl) { ???????if(shortToLong.containsKey(shortUrl)){ ???????????return shortToLong.get(shortUrl); ???????} ???????return ""; ???}}// Your Codec object will be instantiated and called as such:// Codec codec = new Codec();// codec.decode(codec.encode(url));LeetCode - Encode and Decode TinyURL
原文地址：https://www.cnblogs.com/incrediblechangshuo/p/9712302.html