一个完整的非递归版本，照顾到不同的场景:

视图控制器正在呈现另一个视图 视图控制器是一个UINavigationController 视图控制器是一个UITabBarController

objective - c

 UIViewController *topViewController = self.window.rootViewController;
 while (true)
 {
     if (topViewController.presentedViewController) {
         topViewController = topViewController.presentedViewController;
     } else if ([topViewController isKindOfClass:[UINavigationController class]]) {
         UINavigationController *nav = (UINavigationController *)topViewController;
         topViewController = nav.topViewController;
     } else if ([topViewController isKindOfClass:[UITabBarController class]]) {
         UITabBarController *tab = (UITabBarController *)topViewController;
         topViewController = tab.selectedViewController;
     } else {
         break;
     }
 }

斯威夫特 4+

extension UIWindow {
    func topViewController() -> UIViewController? {
        var top = self.rootViewController
        while true {
            if let presented = top?.presentedViewController {
                top = presented
            } else if let nav = top as? UINavigationController {
                top = nav.visibleViewController
            } else if let tab = top as? UITabBarController {
                top = tab.selectedViewController
            } else {
                break
            }
        }
        return top
    }
}

如果根控制器是一个导航控制器，找到顶部可见控制器的正确方法是:

UIViewController *rootVC = [[UIApplication sharedApplication] keyWindow].rootViewController;
if ([rootVC respondsToSelector:@selector(visibleViewController)])
{
    UIViewController *topVC = [(UINavigationController *)rootVC visibleViewController];
    // do your thing with topVC
}

以下是UINavigationController.h的节选:

@property(nonatomic,readonly,retain) UIViewController *topViewController; // The top view controller on the stack.
@property(nonatomic,readonly,retain) UIViewController *visibleViewController; // Return modal view controller if it exists. Otherwise the top view controller.

扩展@Eric的回答，你需要小心，keyWindow实际上是你想要的窗口。例如，如果您试图在点击警报视图中的某些内容后使用此方法，keyWindow实际上将是警报的窗口，这无疑会给您带来问题。这发生在我在野外通过警报处理深度链接时，并导致SIGABRTs没有堆栈跟踪。要调试的婊子。

下面是我现在使用的代码:

- (UIViewController *)getTopMostViewController {
    UIWindow *topWindow = [UIApplication sharedApplication].keyWindow;
    if (topWindow.windowLevel != UIWindowLevelNormal) {
        NSArray *windows = [UIApplication sharedApplication].windows;
        for(topWindow in windows)
        {
            if (topWindow.windowLevel == UIWindowLevelNormal)
                break;
        }
    }

    UIViewController *topViewController = topWindow.rootViewController;

    while (topViewController.presentedViewController) {
        topViewController = topViewController.presentedViewController;
    }

    return topViewController;
}

你可以把这个和你喜欢的从这个问题的其他答案中检索顶视图控制器的方法混合在一起。

Swift替代解决方案:

static func topMostController() -> UIViewController {
    var topController = UIApplication.sharedApplication().keyWindow?.rootViewController
    while (topController?.presentedViewController != nil) {
        topController = topController?.presentedViewController
    }

    return topController!
}

以下是我的看法。感谢@Stakenborg指出了跳过UIAlertView作为最顶层控制器的方法

-(UIWindow *) returnWindowWithWindowLevelNormal
{
    NSArray *windows = [UIApplication sharedApplication].windows;
    for(UIWindow *topWindow in windows)
    {
        if (topWindow.windowLevel == UIWindowLevelNormal)
            return topWindow;
    }
    return [UIApplication sharedApplication].keyWindow;
}

-(UIViewController *) getTopMostController
{
    UIWindow *topWindow = [UIApplication sharedApplication].keyWindow;
    if (topWindow.windowLevel != UIWindowLevelNormal)
    {
        topWindow = [self returnWindowWithWindowLevelNormal];
    }

    UIViewController *topController = topWindow.rootViewController;
    if(topController == nil)
    {
        topWindow = [UIApplication sharedApplication].delegate.window;
        if (topWindow.windowLevel != UIWindowLevelNormal)
        {
            topWindow = [self returnWindowWithWindowLevelNormal];
        }
        topController = topWindow.rootViewController;
    }

    while(topController.presentedViewController)
    {
        topController = topController.presentedViewController;
    }

    if([topController isKindOfClass:[UINavigationController class]])
    {
        UINavigationController *nav = (UINavigationController*)topController;
        topController = [nav.viewControllers lastObject];

        while(topController.presentedViewController)
        {
            topController = topController.presentedViewController;
        }
    }

    return topController;
}

这是对Eric的回答的改进:

UIViewController *_topMostController(UIViewController *cont) {
    UIViewController *topController = cont;

    while (topController.presentedViewController) {
        topController = topController.presentedViewController;
    }

    if ([topController isKindOfClass:[UINavigationController class]]) {
        UIViewController *visible = ((UINavigationController *)topController).visibleViewController;
        if (visible) {
            topController = visible;
        }
    }

    return (topController != cont ? topController : nil);
}

UIViewController *topMostController() {
    UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;

    UIViewController *next = nil;

    while ((next = _topMostController(topController)) != nil) {
        topController = next;
    }

    return topController;
}

UIViewController *cont是一个辅助函数。

现在你所需要做的就是调用topMostController()和最顶端的UIViewController应该被返回!