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[LeetCode] Encode and Decode TinyURL 编码和解码小URL地址

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.

这道题让我们编码和解码精简URL地址，这其实很有用，因为有的链接地址特别的长，就很烦，如果能精简成固定的长度，就很清爽。最简单的一种编码就是用个计数器，当前是第几个存入的url就编码成几，然后解码的时候也能根据数字来找到原来的url，参见代码如下：

解法一：

class Solution {
public:

    // Encodes a URL to a shortened URL.
    string encode(string longUrl) {
        url.push_back(longUrl);
        return "http://tinyurl.com/" + to_string(url.size() - 1);
    }

    // Decodes a shortened URL to its original URL.
    string decode(string shortUrl) {
        auto pos = shortUrl.find_last_of("/");
        return url[stoi(shortUrl.substr(pos + 1))];
    }
    
private:
    vector<string> url;
};

上面这种方法虽然简单，但是缺点却很多，首先，如果接受到多次同一url地址，仍然会当错不同的url来处理。当然这个缺点可以通过将vector换成哈希表，每次先查找url是否已经存在。虽然这个缺点可以克服掉，但是由于是用计数器编码，那么当前服务器存了多少url就曝露出来了，也许会有安全隐患。而且计数器编码另一个缺点就是数字会不断的增大，那么编码的长度也就不是确定的了。而题目中明确推荐了使用六位随机字符来编码，那么我们只要在所有大小写字母和数字中随机产生6个字符就可以了，我们用哈希表建立6位字符和url之间的映射，如果随机生成的字符之前已经存在了，我们就继续随机生成新的字符串，直到生成了之前没有的字符串为止。下面的代码中使用了两个哈希表，目的是为了建立六位随机字符串和url之间的相互映射，这样进来大量的相同url时，就不用生成新的随机字符串了。当然，不加这个功能也能通过OJ，这道题的OJ基本上是形同虚设，两个函数分别直接返回参数字符串也能通过OJ，囧～

解法二：

class Solution {
public:
    Solution() {
        dict = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
        short2long.clear();
        long2short.clear();
        srand(time(NULL));
    }

    // Encodes a URL to a shortened URL.
    string encode(string longUrl) {
        if (long2short.count(longUrl)) {
            return "http://tinyurl.com/" + long2short[longUrl];
        }
        int idx = 0, val = 0;
        string randStr;
        for (int i = 0; i < 6; ++i) randStr.push_back(dict[rand() % 62]);
        while (short2long.count(randStr)) {
            randStr[idx] = dict[rand() % 62];
            idx = (idx + 1) % 5;
        }
        short2long[randStr] = longUrl;
        long2short[longUrl] = randStr;
        return "http://tinyurl.com/" + randStr;
    }

    // Decodes a shortened URL to its original URL.
    string decode(string shortUrl) {
        string randStr = shortUrl.substr(shortUrl.find_last_of("/") + 1);
        return short2long.count(randStr) ? short2long[randStr] : shortUrl;
    }
    
private:
    unordered_map<string, string> short2long, long2short;
    string dict;
};

参考资料：

https://discuss.leetcode.com/topic/81637/two-solutions-and-thoughts/2

https://discuss.leetcode.com/topic/81736/c-solution-using-random-just-for-fun

LeetCode All in One 题目讲解汇总(持续更新中...)

文章来自：http://www.cnblogs.com/grandyang/p/6562209.html