题目
求两个视图的最近公共父视图
暴力解法
采用的是两个 while 循环嵌套迭代来查询最近的公共父视图, 时间复杂度 O(n²), 详见 Masonry 的实现：
- (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;
}
空间换时间的优化解法
- (UIView *)sqi_closestCommonSuperView:(UIView *)view {UIView *closestCommonSuperview = nil;// 用于存放fooView的所有父视图NSMutableSet *cacheViews = [NSMutableSet set];UIView *fooViewSuperview = view;while (fooViewSuperview) {[cacheViews addObject:fooViewSuperview];fooViewSuperview = fooViewSuperview.superview;}UIView *barViewSuperview = self;while (!closestCommonSuperview && barViewSuperview) {// containsObject方法的时间复杂度为1if ([cacheViews containsObject:barViewSuperview]) {closestCommonSuperview = barViewSuperview;}barViewSuperview = barViewSuperview.superview;}return closestCommonSuperview;
}
)
