在iOS中反转字符串

我正在写一段代码，它应该反转一个字符串。当我运行以下代码时，它会抛出一个错误：

- (NSString*) reversingName:(NSString *)myNameText
{
   NSString *result;
   int len = [myNameText length];
   NSMutableString *reverseName = [[NSMutableString alloc] initWithCapacity:len];
   for(int i=len;i>0;i--)
   {
     [reverseName appendFormat:[NSString stringWithFormat:@"%c",[myNameText characterAtIndex:i]]];
   }
   result = reverseName;
   return result;
}

---

环线的

应如下所示：

for(int i=len-1;i>=0;i--)

所以你的方法应该是

- (NSString*) reversingName:(NSString *)myNameText
{
    int len = [myNameText length];
    NSMutableString *reverseName = [[NSMutableString alloc] initWithCapacity:len];
    for(int i=len-1;i>=0;i--)
    {
        [reverseName appendString:[NSString stringWithFormat:@"%c",[myNameText characterAtIndex:i]]];
    }
    return [reverseName autorelease];
}

请尝试以下示例代码：

NSString *name = @"abcdefghi" ;
    int len = [name length];
    NSMutableString *reverseName = [[NSMutableString alloc] initWithCapacity:len];

    for(int i=len-1;i>=0;i--)
    {
        [reverseName appendFormat:[NSString stringWithFormat:@"%c",[name characterAtIndex:i]]];

    }

    NSLog(@"%@",reverseName);

我想如果有人感兴趣的话，我会推出另一个版本。。就我个人而言，我喜欢使用NSMutableString的更干净的方法，但如果性能是最高优先级，则此方法会更快：

- (NSString *)reverseString:(NSString *)input {
    NSUInteger len = [input length];
    unichar *buffer = malloc(len * sizeof(unichar));
    if (buffer == nil) return nil; // error!
    [input getCharacters:buffer];

    // reverse string; only need to loop through first half
    for (NSUInteger stPos=0, endPos=len-1; stPos < len/2; stPos++, endPos--) {
        unichar temp = buffer[stPos];
        buffer[stPos] = buffer[endPos];
        buffer[endPos] = temp;
    }

    return [[NSString alloc] initWithCharactersNoCopy:buffer length:len freeWhenDone:YES];
}

（注意：我还没有足够的声望点数来投票或评论答案，所以如果有人能为我投票，我将不胜感激。我已经阅读了很长时间，但现在想开始贡献更多！）
在Swift 2.0中反转字符串：

let string = "This is a test string."
let characters = string.characters
let reversedCharacters = characters.reverse()
let reversedString = String(reversedCharacters)

String("This is a test string.".characters.reverse())

String(Array("This is a test string.").reverse())

捷径：

let string = "This is a test string."
let characters = string.characters
let reversedCharacters = characters.reverse()
let reversedString = String(reversedCharacters)

String("This is a test string.".characters.reverse())

String(Array("This is a test string.").reverse())

或

捷径：

let string = "This is a test string."
let characters = string.characters
let reversedCharacters = characters.reverse()
let reversedString = String(reversedCharacters)

String("This is a test string.".characters.reverse())

String(Array("This is a test string.").reverse())

[reverseName appendFormat:@“%c”，[name characterAtIndex:i]]
已经足够了。您甚至可以使用XOR操作来提高速度（而不是每次分配一个临时变量）