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NSMutableString *str =@"abcdefg123";

我想将每个字符随机到一个像这样@“f1ad2g3be2”的新字符串。

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2 回答 2

3

非常简单。首先,您必须将字符分解为要使用的数组。然后你多次交换字母 X,我选择这样做,所以每个字符都会被交换

NSString *str =@"abcdefg123";
int length = str.length;
NSMutableArray *letters = [[NSMutableArray alloc] init];
for (int i = 0; i< length; i++) {
    NSString *letter = [NSString stringWithFormat:@"%c", [str characterAtIndex:i]];
    [letters addObject:letter];
}

for (int i = 0; i<length; i++) {
    int value = arc4random() % (length-1);
    NSLog(@"Value is : %i", value);
    [letters exchangeObjectAtIndex:i withObjectAtIndex:value];
}

NSString *results = [letters componentsJoinedByString:@""];
NSLog(@"The string before : %@", str);
NSLog(@"This is the string now : %@", results);
于 2012-10-19T11:38:37.627 回答
3
NSMutableString *str1 = [[NSMutableString alloc]initWithString:str];
NSMutableString *str2 = [[NSMutableString alloc] init];
while ([str1 length] > 0) {
    int i = arc4random() % [str1 length];
    NSRange range = NSMakeRange(i,1);
    NSString *sub = [str1 substringWithRange:range];
    [str2 appendString:sub];
    [str1 replaceOccurrencesOfString:sub withString:@"" options:nil range:range];
}

[str1 release];

str2 是你想要的

于 2012-10-19T11:24:27.327 回答