Question

我想要随机化字母表中所有字符的字符串。现在，我创建了一个包含26个字符的可变数组，使用exchangeObjectAtIndex：方法对它们进行随机播放，然后将每个字符添加到我返回的字符串中。

必须有更好的方法来做到这一点。这是我的代码：

- (NSString *)shuffledAlphabet {
    NSMutableArray * shuffledAlphabet = [NSMutableArray arrayWithArray:@[@"A",@"B",@"C",@"D",@"E",@"F",@"G",@"H",@"I",@"J",@"K",@"L",@"M",@"N",@"O",@"P",@"Q",@"R",@"S",@"T",@"U",@"V",@"W",@"X",@"Y",@"Z"]];

    for (NSUInteger i = 0; i < [shuffledAlphabet count]; ++i) {
        // Select a random element between i and end of array to swap with.
        int nElements = [shuffledAlphabet count] - i;
        int n = (random() % nElements) + i;
        [shuffledAlphabet exchangeObjectAtIndex:i withObjectAtIndex:n];
    }

    NSString *string = [[NSString alloc] init];
    for (NSString *letter in shuffledAlphabet) {
        string = [NSString stringWithFormat:@"%@%@",string,letter];
    }

    return string;
}

Answer 1

这是一个高效的Fisher-Yates shuffle，适合您的用例：

- (NSString *)shuffledAlphabet {
    NSString *alphabet = @"ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    // Get the characters into a C array for efficient shuffling
    NSUInteger numberOfCharacters = [alphabet length];
    unichar *characters = calloc(numberOfCharacters, sizeof(unichar));
    [alphabet getCharacters:characters range:NSMakeRange(0, numberOfCharacters)];

    // Perform a Fisher-Yates shuffle
    for (NSUInteger i = 0; i < numberOfCharacters; ++i) {
        NSUInteger j = (arc4random_uniform(numberOfCharacters - i) + i);
        unichar c = characters[i];
        characters[i] = characters[j];
        characters[j] = c;
    }

    // Turn the result back into a string
    NSString *result = [NSString stringWithCharacters:characters length:numberOfCharacters];
    free(characters);
    return result;
}

Answer 2

这是执行正确改组的字母表生成的更有效方法。

- (NSString *)shuffledAlphabet
{
    const NSUInteger length = 'Z' - 'A' + 1;
    unichar alphabet[length];
    alphabet[0] = 'A';

    for ( NSUInteger i = 1; i < length; i++ )
    {
        NSUInteger j = arc4random_uniform((uint32_t)i + 1);
        alphabet[i] = alphabet[j];
        alphabet[j] = 'A' + i;
    }
    return [NSString stringWithCharacters:alphabet length:length];
}

它使用Fischer Yates shuffle的“由内而外”版本，并通过使用arc4random_uniform生成伪随机数来避免模块偏差。此外，它需要单个分配，因为所有排列都在临时缓冲区中执行。

Answer 3

Generating random numbers in Objective-C这有帮助吗？ *生成随机数 *除以26并提醒 *索引数组[提醒]

生成随机字母串的有效方法？

3 个答案: