Objective-C 中的 Base 62 转换

Question

我花了太多时间试图找到 Objective-C 的 base 62 转换的实现。我确信这是一个糟糕的例子，必须有一种优雅、超高效的方法来做到这一点，但这是有效的，请编辑或回答以改进它！但我想帮助搜索此内容的人找到有用的东西。对于 Objective-C 实现，似乎找不到任何特定的东西。

@implementation Base62Converter

+(int)decode:(NSString*)string
{
    int num = 0;
    NSString * alphabet = @"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";

    for (int i = 0, len = [string length]; i < len; i++)
    {
        NSRange range = [alphabet rangeOfString:[string substringWithRange:NSMakeRange(i,1)]];
        num = num * 62 + range.location;
    }

    return num;
}

+(NSString*)encode:(int)num
{
    NSString * alphabet = @"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
    NSMutableString * precursor = [NSMutableString stringWithCapacity:3];

    while (num > 0)
    {
        [precursor appendString:[alphabet substringWithRange:NSMakeRange( num % 62, 1 )]];
        num /= 62;
    }

    // http://stackoverflow.com/questions/6720191/reverse-nsstring-text
    NSMutableString *reversedString = [NSMutableString stringWithCapacity:[precursor length]];

    [precursor enumerateSubstringsInRange:NSMakeRange(0,[precursor length])
                             options:(NSStringEnumerationReverse |NSStringEnumerationByComposedCharacterSequences)
                          usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
                              [reversedString appendString:substring];
                          }];
    return reversedString;
}

@end

Answer 1

你的代码没问题。如果有的话，让它更通用。这是任何基础（相同代码）的递归版本：

#import <Foundation/Foundation.h>

@interface BaseConversion : NSObject
+(NSString*) formatNumber:(NSUInteger)n toBase:(NSUInteger)base;
+(NSString*) formatNumber:(NSUInteger)n usingAlphabet:(NSString*)alphabet;
@end
 
@implementation BaseConversion

// Uses the alphabet length as base.
+(NSString*) formatNumber:(NSUInteger)n usingAlphabet:(NSString*)alphabet
{
    NSUInteger base = [alphabet length];
    if (n<base){
        // direct conversion
        NSRange range = NSMakeRange(n, 1);
        return [alphabet substringWithRange:range];
    } else {
        return [NSString stringWithFormat:@"%@%@",

                // Get the number minus the last digit and do a recursive call.
                // Note that division between integer drops the decimals, eg: 769/10 = 76
                [self formatNumber:n/base usingAlphabet:alphabet],

                // Get the last digit and perform direct conversion with the result.
                [alphabet substringWithRange:NSMakeRange(n%base, 1)]];
    }
}

+(NSString*) formatNumber:(NSUInteger)n toBase:(NSUInteger)base 
{
    NSString *alphabet = @"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; // 62 digits
    NSAssert([alphabet length]>=base,@"Not enough characters. Use base %ld or lower.",(unsigned long)[alphabet length]);
    return [self formatNumber:n usingAlphabet:[alphabet substringWithRange:NSMakeRange (0, base)]];
}
 
@end
 
int main(int argc, char *argv[]) {
    @autoreleasepool {
        NSLog(@"%@",[BaseConversion formatNumber:3735928559 toBase:16]); // deadbeef
        return EXIT_SUCCESS;
    }
}

Swift 版本：https://gist.github.com/janodev/9fb63a408251e66d7f1756f2537bc4b6

Answer 2

你可以改进你的

encode

方法，这样就不需要反转最后的字符串：

+ (NSString *)encode:(NSUInteger)num
{
    NSString *alphabet = @"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
    NSUInteger base = [alphabet length];
    NSMutableString *result = [NSMutableString string];
    while (num > 0) {
        NSString *digit = [alphabet substringWithRange:NSMakeRange(num % base, 1)];
        [result insertString:digit atIndex:0];
        num /= base;
    }
    return result;
}

当然，这也可以推广到任意基数或字母表，正如@Jano 在他的回答中所建议的那样。

请注意，此方法（以及您原来的

encode

方法）为

num = 0

返回一个空字符串，因此您可能需要单独考虑这种情况（或者只需将

while (num > 0) { ... }

替换为

do { ... } while (num > 0)

.

---

为了提高效率，可以完全避免所有中间

NSString

对象，并使用纯 C 字符串：

+ (NSString *)encode:(NSUInteger)num
{
    static const char *alphabet = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
    NSUInteger base = 62;

    char result[20]; // sufficient room to encode 2^64 in Base-62
    char *p = result + sizeof(result);

    *--p = 0; // NULL termination
    while (num > 0) {
        *--p = alphabet[num % base];
        num /= base;
    }
    return [NSString stringWithUTF8String:p];
}