生成随机字母字符串的有效方法？

我想要一个随机字母表中所有字符的字符串。现在，我创建一个包含 26 个字符的可变数组，使用 ExchangeObjectAtIndex: 方法对它们进行打乱，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;
}

这里有一个高效的费舍尔-耶茨洗牌 http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_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;
}