查找两个view最近公共祖先那段逻辑算法复杂度是O(n²)，可以优化

[hopestar90]

New Issue Checklist

Issue Info

如下代码:

• (instancetype)mas_closestCommonSuperview:(MAS_VIEW *)view {
  MAS_VIEW *closestCommonSuperview = nil;

  MAS_VIEW *secondViewSuperview = view;
  while (!closestCommonSuperview && secondViewSuperview) {
  MAS_VIEW *firstViewSuperview = self;
  while (!closestCommonSuperview && firstViewSuperview) {
  if (secondViewSuperview == firstViewSuperview) {
  closestCommonSuperview = secondViewSuperview;
  }
  firstViewSuperview = firstViewSuperview.superview;
  }
  secondViewSuperview = secondViewSuperview.superview;
  }
  return closestCommonSuperview;
  }

上述代码意图为寻找两个View最近的公共父View，时间复杂度是O(n²)，其实可以优化成O(n)。

将两个view作为两个链表的头结点，最顶层的superview作为尾结点，这样问题就转化为寻找两个链表的第一个相交节点的问题，从而变为一个复杂度为O(n)的问题。不知道这么考虑是否是正确的？

Issue Description

⚠️ Replace this with the description of your issue. ⚠️

[Jake-Chiu]

有想法是好的，你可以把你的优化方案写出来，验证一下不就ok了吗

[cntrump]

这样？

- (__kindof MAS_VIEW *)mas_closestCommonSuperview:(MAS_VIEW *)view {
    MAS_VIEW *closestCommonSuperview = self;
    MAS_VIEW *secondViewSuperview = view;

    while (closestCommonSuperview != secondViewSuperview) {
        closestCommonSuperview = closestCommonSuperview == nil ? view : closestCommonSuperview.superview;
        secondViewSuperview = secondViewSuperview == nil ? self : secondViewSuperview.superview;
    }

    return closestCommonSuperview;
}

[hopestar90]

这么写感觉在某些case会陷入死循环，如果self的superview是nil,view也是nil的情况

[cntrump]

这么写感觉在某些case会陷入死循环，如果self的superview是nil,view也是nil的情况

返回 nil。
理论上不会死循环，因为不会形成环。
这里有别人做的动画演示。 https://zhuanlan.zhihu.com/p/357611418

[hopestar90]

我给一个写法~ 但可能没有那么优雅:

MAS_VIEW *firstView = self;
MAS_VIEW *secondView = view;
NSInteger firstLength = 0;
NSInteger secondLength = 0;
while (firstView) {
firstLength++;
firstView = firstView.superview;
}
while (secondView) {
secondLength++;
secondView = secondView.superview;
}
NSInteger diff = 0;
if (firstLength > secondLength) {
diff = firstLength - secondLength;
firstView = self;
secondView = view;
} else {
diff = secondLength - firstLength;
firstView = view;
secondView = self;
}

while (diff > 0) {
    diff--;
    firstView = firstView.superview;
}
while (firstView && secondView && firstView != secondView) {
    firstView = firstView.superview;
    secondView = secondView.superview;
}
if (firstView != secondView) {
    return nil;
} else {
    return firstView;
}

[hopestar90]

这样？

- (__kindof MAS_VIEW *)mas_closestCommonSuperview:(MAS_VIEW *)view {
    MAS_VIEW *closestCommonSuperview = self;
    MAS_VIEW *secondViewSuperview = view;

    while (closestCommonSuperview != secondViewSuperview) {
        closestCommonSuperview = closestCommonSuperview == nil ? view : closestCommonSuperview.superview;
        secondViewSuperview = secondViewSuperview == nil ? self : secondViewSuperview.superview;
    }

    return closestCommonSuperview;
}

刚又分析了下 看了下 这样确实可以返回正确的结果

[OPTJoker]

楼上都搞的太复杂了，只需要几个简单的if-else就可以优化95%的遍历次数。
毕竟firstView与secondView，在大多数场景下都是父子关系 or 兄弟关系。

优化代码如下：

#pragma mark - heirachy

- (instancetype)mas_closestCommonSuperview:(MAS_VIEW *)view {
    MAS_VIEW *closestCommonSuperview = nil;
    
    // 一般约束关联的两个view要么是父子关系，要么是兄弟关系，碰运气可以大大优化链表遍历的时间
    closestCommonSuperview = [self take_a_chance:view];
    if (closestCommonSuperview)
        return closestCommonSuperview;

    MAS_VIEW *secondViewSuperview = view;
    while (!closestCommonSuperview && secondViewSuperview) {
        MAS_VIEW *firstViewSuperview = self;
        while (!closestCommonSuperview && firstViewSuperview) {
            if (secondViewSuperview == firstViewSuperview) {
                closestCommonSuperview = secondViewSuperview;
            }
            firstViewSuperview = firstViewSuperview.superview;
        }
        secondViewSuperview = secondViewSuperview.superview;
    }
    return closestCommonSuperview;
}


/// 碰运气 快速找到common super view
- (MAS_VIEW *)take_a_chance:(MAS_VIEW *)secondView {
    if (!secondView) return nil;
    if (self == secondView) return self;
    
    if (secondView.superview == self.superview && self.superview) {
        return self.superview; // 兄弟关系
    } else if (self.superview == secondView) {
        return self.superview; // first 是 second 的子view
    } else if (self == secondView.superview) {
        return secondView.superview; // first 是 second 的父view
    }
    return nil;
}

优化结果如下：（某日活百万app启动，到首页不再layout）

优化前：
查找公共viev:323次
查找父view遍历：1656次

优化后：
查找公共view：10次
查找父view遍历：84次

[cntrump]

楼上都搞的太复杂了，只需要几个简单的if-else就可以优化95%的遍历次数。 毕竟firstView与secondView，在大多数场景下都是父子关系 or 兄弟关系。

优化代码如下：

#pragma mark - heirachy

- (instancetype)mas_closestCommonSuperview:(MAS_VIEW *)view {
    MAS_VIEW *closestCommonSuperview = nil;
    
    // 一般约束关联的两个view要么是父子关系，要么是兄弟关系，碰运气可以大大优化链表遍历的时间
    closestCommonSuperview = [self take_a_chance:view];
    if (closestCommonSuperview)
        return closestCommonSuperview;

    MAS_VIEW *secondViewSuperview = view;
    while (!closestCommonSuperview && secondViewSuperview) {
        MAS_VIEW *firstViewSuperview = self;
        while (!closestCommonSuperview && firstViewSuperview) {
            if (secondViewSuperview == firstViewSuperview) {
                closestCommonSuperview = secondViewSuperview;
            }
            firstViewSuperview = firstViewSuperview.superview;
        }
        secondViewSuperview = secondViewSuperview.superview;
    }
    return closestCommonSuperview;
}


/// 碰运气 快速找到common super view
- (MAS_VIEW *)take_a_chance:(MAS_VIEW *)secondView {
    if (!secondView) return nil;
    if (self == secondView) return self;
    
    if (secondView.superview == self.superview && self.superview) {
        return self.superview; // 兄弟关系
    } else if (self.superview == secondView) {
        return self.superview; // first 是 second 的子view
    } else if (self == secondView.superview) {
        return secondView.superview; // first 是 second 的父view
    }
    return nil;
}

优化结果如下：（某日活百万app启动，到首页不再layout）

优化前： 查找公共viev:323次 查找父view遍历：1656次

优化后： 查找公共view：10次 查找父view遍历：84次

LGTM，搬走抄到我的 Fork 里