如何让UILabel点击像UIButton一样打开URL

import UIKit

class ViewController: UIViewController {
    override func viewDidLoad() {
        super.viewDidLoad()
        
        let label = UILabel(frame: CGRect(x: 100, y: 100, width: 200, height: 30))
        label.text = "点击打开URL"
        label.isUserInteractionEnabled = true
        
        let tapGesture = UITapGestureRecognizer(target: self, action: #selector(labelTapped))
        label.addGestureRecognizer(tapGesture)
        
        view.addSubview(label)
    }
    
    @objc func labelTapped() {
        guard let url = URL(string: "https://www.example.com") else { return }
        UIApplication.shared.open(url)
    }
}