Description
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.
分析
题目的意思是:给定一个url,编码乘TinyURL,并且保证TinyURL能够解码回来。
这道题我也不怎么会,后面发现别人做得也是很简单的,encode阶段,如果该url已经编码了,则直接返回已经编码的url,如果没有,则生成随机数,然后判断随机数是否已经生成,如果生成了,则重新生成,然后存入url 的short long和long short的键值对
代码
class Solution {
private:
unordered_map<string,string> short2long,long2short;
string dict;
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)){
randStr[idx]=dict[rand()%62];
idx=(idx+1)%5;
}
short2long[randStr]=longUrl;
long2short[longUrl]=randStr;
return "http://tinyurl.com/" + long2short[longUrl];
}
// 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;
}
};
// Your Solution object will be instantiated and called as such:
// Solution solution;
// solution.decode(solution.encode(url));