Can you solve this real interview question? Print in Order - Suppose we have a class: public class Foo { public void first() { print("first"); } public void second() { print("second"); } public void third() { print("third"); } } The same instan
printSecond(); m1.unlock(); m2.unlock(); } void third(function<void()> printThird) { m2.lock(); // printThird() outputs "third". Do not change or remove this line. printThird(); m2.unlock(); } }; void printFirst() { cout << "first"; } void printSecond() { cout <<...
We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seems to imply the ordering. The input format you see is mainly to ensure our tests' comprehensiveness. 我们提供了一个类: public class Foo { public void one() { print("one")...
解法思路：首先，引入三个无缓冲channel，并将其集合为一个slice。使用PrintInOrder()函数按指定顺序启动三个goroutine。三个goroutine中，只有First()可以自由执行打印，而Second()和Third()则需等待各自的sync信号，以此确保First先输出。接着，当Second执行完成后，由sync[1]传递控制权，依次类推至Thir...
本文旨在解析 LeetCode Concurrency（并发）挑战中的"Print In Order"问题，利用Go语言进行解答。探讨此题的考核点，解题策略以及代码实现，并解答读者提问。问题考核点在于，设计三个goroutine（First(), Second(), Third()），确保不论它们的执行顺序如何，输出的文本顺序必须保持一致，不受执行顺序影响...
来源:力扣(LeetCode)链接:https://leetcode-cn.com/problems/print-in-order 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 题解 使用juc包的CountDownLatch 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classFoo{
来源:https://leetcode-cn.com/problems/print-in-order/ 方案一:CountDownLatch 由于在该场景中,每个方法只会被调用一次,因此用CountDownLatch来控制执行阶段是最自然的思路: CountDownLatch stage1 = new CountDownLatch(1); CountDownLatch stage2 = new CountDownLatch(1); public void first(Runnable print...
{LeetCode144 l=newLeetCode144();//构造二叉树TreeNode root=newTreeNode(1);TreeNode node1=newTreeNode(2);TreeNode node2=newTreeNode(3);root.left=node1;node1.right=node2;//二叉树先序遍历List<Integer>nodes=l.preorderTraversal(root);nodes.stream().forEach(n->{System.out.print(n);})...
(10): yield i*2 for i in foo(): print(i) l = [i for i in range(100000)] # 一次性生成好 l = (i for i in range(100000)) # 需要对时候再去计算 # 匿名函数 x = lambda a : a + 10 print(x(5)) sum = lambda arg1, arg2: arg1 + arg2 print ("相加后的值为 : ", ...
print-binary-tree print-foobar-alternately print-immutable-linked-list-in-reverse print-in-order print-zero-even-odd prison-cells-after-n-days product-of-array-except-self product-of-the-last-k-numbers promise-pool promise-time-limit qIsx9U qi-wang-ge-shu-tong-ji qing-wa...