535. Encode and Decode TinyURL

问题描述：

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.

解题思路：

这道题看到要将长的转换成短的，我能想到hashmap，但是对碰撞问题我是没有什么好的解决方案。

看了Grandyang的分析。

这里使用了rand()函数来随机生成并从字典里获取字符。

生成完了6位字符串后，需要检查是否发生碰撞，若发生碰撞则重新生成。

用了两个hashmap来存储：

short2long：主要用于转换

long2short：主要用于查重

代码：

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];
        string randStr;
        for(int i = 0; i < 6; i++){
            randStr.push_back(dict[rand() % 62]);
        }
        
        int idx = 0;
        while(short2long.count(randStr)){
            idx %= 6;
            randStr[idx] = dict[rand() % 62];
        }
        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;
};

// Your Solution object will be instantiated and called as such:
// Solution solution;
// solution.decode(solution.encode(url));