535. Encode and Decode TinyURL

題目

https://leetcode.com/problems/encode-and-decode-tinyurl/

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.

解答

這道題是讓我們設計一個短url的編碼方式。

其實可以簡單粗暴：

public List<String> data = new ArrayList<>();

public String encode(String longUrl) {
    int id = data.size();
    data.add(longUrl);

    return id + "";
}

// Decodes a shortened URL to its original URL.
public String decode(String shortUrl) {
    int id = Integer.valueOf(shortUrl);
    return data.get(id);
}

不過這樣不就沒有符合出題人的意圖了嘛，我搜了一下如何shorter url，發現原來是個進制轉換的過程。
https://stackoverflow.com/questions/742013/how-do-i-create-a-url-shortener

然後我就嘗試寫了一下，stackoverflow上寫的是62進制，其實你只要設計一個字母表，可以是任意進制的

public List<String> data = new ArrayList<>();

// Encodes a URL to a shortened URL.
public String encode(String longUrl) {
    int id = data.size();
    data.add(longUrl);
    return convert(id);
}

// Decodes a shortened URL to its original URL.
public String decode(String shortUrl) {
    return data.get(convert(shortUrl));
}

// 字母表，隨便設計，只要字符不重複就可以
public static final String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*_+-=~`";

public static String convert(int number) {
    if (number == 0) {
        return alphabet.charAt(0) + "";
    }
    StringBuilder sb = new StringBuilder();
    while (number > 0) {
        sb.append(alphabet.charAt(number % alphabet.length()));
        number /= alphabet.length();
    }
    return sb.reverse().toString();
}

public static int convert(String code) {
    int num = 0;
    for (int i = 0; i < code.length(); i++) {
        num += alphabet.indexOf(code.charAt(i)) * Math.pow(alphabet.length(), code.length() - i - 1);
    }
    return num;
}

// 這裏是測試代碼
public static void main(String[] args) {
    for (int i = 0; i < 10000; i++) {
        String code = convert(i);
        System.out.println("num=" + i + " code=" + code + " back=" + convert(code));
    }
}

爲了驗證我寫的進制轉換方法是正確的，我寫了個測試方法來檢測看對不對。看起來是對的。

num=0 code=a back=0
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