如何将长度为3的Base62字符串反向转换为原始整数？

Absolutely! You can definitely reverse this Base62 encoding to get back your original integer—since it's a fixed-length, one-to-one mapping between integers and 3-character strings, the process is straightforward. Here's how to make it work, tailored to your example where "zzz" maps back to 238328.

How the Reverse Process Works #

Your encoding uses a 62-character set (0-9, A-Z, a-z), where each position in the 3-character string has a weight based on powers of 62:

• The first (leftmost) character has a weight of 62² (3844)
• The second character has a weight of 62¹ (62)
• The third (rightmost) character has a weight of 62⁰ (1)

For a string like "zzz", each 'z' is the 61st index in your CHARACTERS string (since we start counting from 0). The raw Base62 value would be:
61*3844 + 61*62 + 61 = 238327

Since your encoding maps 238328 to "zzz", it’s likely adding 1 to the raw Base62 value during encoding. So we just need to reverse that step in decoding.

Java Implementation for Decoding #

Here’s a complete decoding method that matches your original encoding setup:

private static final String CHARACTERS = "0123456789" + "ABCDEFGHIJKLMNOPQRSTUVWXYZ" + "abcdefghijklmnopqrstuvwxyz";

public static int decodeBase62(String encoded) {
    // Validate input length matches your fixed 3-character requirement
    if (encoded.length() != 3) {
        throw new IllegalArgumentException("Encoded string must be exactly 3 characters long");
    }
    
    int baseValue = 0;
    for (int i = 0; i < encoded.length(); i++) {
        char currentChar = encoded.charAt(i);
        int charIndex = CHARACTERS.indexOf(currentChar);
        
        // Handle invalid characters not present in your encoding set
        if (charIndex == -1) {
            throw new IllegalArgumentException("Invalid character in encoded string: " + currentChar);
        }
        
        // Calculate the contribution of the current character to the total value
        baseValue += charIndex * (int) Math.pow(62, 2 - i);
    }
    
    // Adjust for the +1 offset used in your encoding (since 238327 +1 = 238328)
    return baseValue + 1;
}

Testing the Example #

If you run decodeBase62("zzz"), it will return 238328—exactly the integer you started with.

Key Notes to Avoid Issues #

• Keep the CHARACTERS string identical: The order of characters (digits first, then uppercase, then lowercase) must match exactly what’s used in your encoding code. Changing the order will break the mapping.
• Input validation: Always check that the input string is 3 characters long and only contains valid characters from your set.
• Offset adjustment: If your original encoding doesn’t use a +1 offset (e.g., 238327 maps to "zzz"), just remove the +1 in the return statement.

内容的提问来源于stack exchange，提问作者cubeb