格特鲁德·伯格 (Gertrude Berg) 的《戈德堡一家》(The Goldbergs) 曾是一部大胆描绘犹太家庭并备受喜爱的作品。然而,随后的黑名单事件却彻底摧毁了她留下的宝贵事业。 伯格的节目始于 1929 年的广播,一经播出便大受欢迎。1949 年,该节目被搬上电视荧屏。/ Getty 1954 年 5 月 9 日,在 CBS 游戏节目《What's My Line?》的录制现场,当...
Java Coding Tips Use less target.setXX(source.getXX) // when you want to use less target.setXX(source.getXX), same name BeanUtils.copyProperties(source, target) Manipulate Array copy array `java // most efficient(recommend) void System.arraycopy(int[] srcArr, int srcBeginP...
摘要: The ZGC garbage collector is designed for low latency, using Region-based memory layout, concurrent mark-compact algorithms, and innovative techniques like colored pointers and multi-mapping. It dynamically manages memory with small, medium, and large pages, optimizing object placement. ZGC's colored pointers store metadata directly in references, reducing memory barriers and enabling efficient concurrent operations. Despite its 4TB memory limit and lack of generational collection, ZGC outperforms other collectors in latency, achieving sub-10ms pauses while maintaining high throughput.
ZGC 收集器是一款基于 Region 内存布局的,(暂时) 不设分代的,使用了读屏障、染色指针和内存多重映射等技术来实现可并发的标记 - 整理算法的,以低延迟为首要目标的一款垃圾收集器。 ZGC的内存布局 与 Shenandoah 和 G1 一样,ZGC 也采用基于 Region 的堆内存布局,但与它们不同的是,ZGC 的 Page 具有动态性——动态创建和销毁,以及动态的区域容量大小。在 x64 硬件平台下,ZGC 的 Page 可以具有如下图所示的大、中、小三类容量: 小型 Page (Small Page):容量固定为 2 MB,用户放置小于 256 KB 的小对象...
摘要: This text analyzes the "Remove Sub-folders from the Filesystem" LeetCode problem and presents two solution approaches. Method 1 uses a Trie (prefix tree) data structure: define trie nodes, insert all files into the tree, then use DFS traversal to collect parent nodes. Method 2 uses sorting: sort files lexicographically, add the first file to results, then traverse remaining elements to check if each is a subfolder of the last added element. For the Trie approach, time complexity is O(nl) and space complexity is O(nl), where n is array length and l is average folder length. For the Sort approach, time complexity is O(nl⋅logn) due to sorting, and space complexity is O(l) for temporary string operations during prefix comparisons.
Question remove sub folders from the filesystem Analysis Method 1: Trie(Prefix tree) define trie node structure insert all file into tree using DFS traverse to collect parent node Method 2:...
摘要: The problem involves balancing fruit costs between two baskets. If the total count of any fruit cost is odd, balancing is impossible. Excess fruits are listed and sorted. A greedy strategy pairs smallest and largest costs to minimize exchange costs, choosing between direct or indirect swaps based on cost efficiency. Time complexity is O(n log n) due to sorting.
Question rearranging fruits Approach: Greedy Intuition According to the problem, the cost x of any fruit must appear an even number of times across the two fruit baskets; otherwise, it is impossible to evenly distribute th...
摘要: 题目要求计算没有会议的天数。解决方法是先将会议按开始时间排序,然后比较后一个会议的开始时间和前一个会议的结束时间,以确定是否有重叠或连续。时间复杂度为O(n),空间复杂度为O(1)。
Question Count days without meetings Analysis sort meetings array by start time of each meeting compare start time of latter and end time of former Timing 21 min 14 sec Code `java public class Main1369 {...
摘要: The problem involves matching players with trainers by sorting two arrays and using a two-pointer technique. The solution has a time complexity of O(n log n) due to sorting and a space complexity of O(log n). It was completed in 7 minutes and 41 seconds.
Question maximum matching of players with trainers Analysis Sort two arrays Two pointer to record matching position Timing 7 min 41 sec Code `java class Solution { public int matchPlaye...