使用swift中颜色的RGB值或十六进制值更改视图的颜色

我在网上得到了下面的代码，我在我的项目中使用了这段代码来改变背景的颜色

// Creates a UIColor from a Hex string.

func colorWithHexString (hex:String) -> UIColor {
    var cString:String = hex.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).uppercaseString

    if (cString.hasPrefix("#")) {
        cString = (cString as NSString).substringFromIndex(1)
    }

    if (countElements(cString) != 6) {
        return UIColor.grayColor()
    }

    var rString = (cString as NSString).substringToIndex(2)
    var gString = ((cString as NSString).substringFromIndex(2) as NSString).substringToIndex(2)
    var bString = ((cString as NSString).substringFromIndex(4) as NSString).substringToIndex(2)

    var r:CUnsignedInt = 0, g:CUnsignedInt = 0, b:CUnsignedInt = 0;
    NSScanner(string: rString).scanHexInt(&r)
    NSScanner(string: gString).scanHexInt(&g)
    NSScanner(string: bString).scanHexInt(&b)


    return UIColor(red: CGFloat(r) / 255.0, green: CGFloat(g) / 255.0, blue: CGFloat(b) / 255.0, alpha: CGFloat(1))
}

当我按如下方式调用上述函数来更改背景颜色时：

lbl91.backgroundColor = colorWithHexString(hex: 0x209624);

它给了我以下错误：

cannot convert the expression's type '()' to type integerliteralconvertible

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函数期望其参数如下所示：

colorWithHexString("#ff00dd")

或者没有标签

colorWithHexString("ff00dd")

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使用字符串而不是十六进制数作为输入参数；可能重复的