python--leetcode 535. 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.

这一题还是比较有意思的，题目意思就是说输入一个网址，让你用算法缩短这个网址，而且能否复原原网址。

当然这一题可以耍点小聪明就过了，直接return longurl 和return shorturl返回原地址。不过这并非我们刷leetcode的本意。

下面上代码，我一会再解释代码的意思：

class Codec:    import string    letters = string.ascii_letters + string.digits  #abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789    full_tiny = {}    tiny_full = {}    global_counter = 0    def encode(self, longUrl):        """Encodes a URL to a shortened URL.        :type longUrl: str        :rtype: str        """        def decto62(dec):            ans = ""            while 1:                ans = self.letters[dec % 62] + ans                dec //= 62                if not dec:                    break            print('ans:',ans)            return ans        suffix = decto62(self.global_counter)        if longUrl not in self.full_tiny:            self.full_tiny[longUrl] = suffix            self.tiny_full[suffix] = longUrl            self.global_counter += 1        return "http://tinyurl.com/" + suffix    def decode(self, shortUrl):        """Decodes a shortened URL to its original URL.        :type shortUrl: str        :rtype: str        """        idx = shortUrl.split('/')[-1]        print('fulltiny',self.full_tiny,self.tiny_full)        if idx in self.tiny_full:            return self.tiny_full[idx]        else:            return Nonecodec = Codec()url='fffffffffff'print(codec.decode(codec.encode(url)))

这里会输出以下内容：

ans: a
fulltiny {'fffffffffff': 'a'} {'a': 'fffffffffff'}
fffffffffff

解题思路：将输入的长地址与生成的短地址通过字典一一对应。dector62方法就是生成短地址的方法，通过global_counter的递增并取余，来获得短地址。