2023领星Java后端开发笔试

• 一、填空题
• 二、简答题
• • 1.Mysql设计
• 三、编程题
• • 1.倒置字符串
  • • (1).二分法(最优)
    • (2).栈 + StringBuffer
    • (3).利用String的toCharArray方法
    • (4).利用StringBuffer
    • (5).正常
  • 2.两个线程交替运行
  • • (1)方法一：用Synchronize+ wait() + notifyAll()
    • (2)方法二：ReentrantLock+ Condition + await() + signal()
  • 3.数组升序排列

一、填空题

1.一个线程OOM后，其它线程还能运行吗？

答：可以。

2.面向对象的三大特性

答：封装、继承、多态

3.实现线程的方式

答：
1.继承Thread类
2.实现Runnable、Callable接口

4.同一个类多个方法的方法名字相同，参数列表不同、参数个数不同、参数的数据类型不同
不同数据类型的参数顺序不同。这种叫

答:重载

5.( )是操作系统资源分配的基本单位，而( )是处理器任务调度和执行的基本单位。

答：进程\线程

二、简答题

1.Mysql设计

某主页需要展示数据，用Mysql数据库，其中主表上亿数据，剩下2张表上百万数据，设计方案，让查询效率快，并且需要考虑少量的写场景。可以用到中间件。

答：看到一篇文章讲述的很好，这里放出链接
解决方案

三、编程题

1.倒置字符串

牛客：NC103 反转字符串
牛客-反转字符串

(1).二分法(最优)

public String solve6(String str) {int length = str.length();if (length == 0) {return "";}char[] chars = str.toCharArray();for (int i = 0; i < length / 2; i++) {char temp = chars[i];chars[i] = chars[length - 1 - i];chars[length - 1 - i] = temp;}return new String(chars);}

(2).栈 + StringBuffer

public String solve5(String str) {int length = str.length();if (length == 0) {return "";}Stack stack = new Stack<>();for (int i = 0; i < length; i++) {stack.push(str.charAt(i));}StringBuffer stringBuffer = new StringBuffer();for (int i = 0; i < length; i++) {stringBuffer.append(stack.pop());}return stringBuffer.toString();}

(3).利用String的toCharArray方法

public String solve(String str) {int length = str.length();if (length == 0) {return "";}char[] chars = str.toCharArray();String result = "";for (int i = length - 1; i >= 0; i--) {result += chars[i];}return result;}

(4).利用StringBuffer

    public String solve(String str) {int length = str.length();if (length == 0) {return "";}return new StringBuffer(str).reverse().toString();}

(5).正常

public String solve2(String str) {int length = str.length();if (0 == length) {return "";}String result = "";for (int i = length - 1; i >= 0; i--) {result += str.charAt(i);}return result;}

2.两个线程交替运行

(1)方法一：用Synchronize+ wait() + notifyAll()

public class ThreadExercise2 {private static String str1 = "ABCD";private static String str2 = "1234";public static void main(String[] args) throws InterruptedException {Object lock = new Object();MyThreadEx myth1 = new MyThreadEx(lock, str1);MyThreadEx myth2 = new MyThreadEx(lock, str2);myth1.start();myth2.start();myth1.join();myth2.join();}
}class MyThreadEx extends Thread {// lock是2个线程需要抢占的变量private Object lock;private String str;public MyThreadEx(Object lock, String str) {this.lock = lock;this.str = str;}@Overridepublic void run() {int length = str.length();// 用synchronized关键字当互斥锁，锁住该对象synchronized (this.lock) {for (int i = 0; i < length; i++) {System.out.println("线程" + Thread.currentThread().getId() + " : " + str.charAt(i));// 这里注意notify会解除阻塞lock.notify();try {// wait与synchronized一起使用会线程阻塞lock.wait();} catch (InterruptedException e) {e.printStackTrace();}}}}
}

(2)方法二：ReentrantLock+ Condition + await() + signal()

public class ThreadExercise3 {private static String str1 = "ABCD";private static String str2 = "1234";private static ReentrantLock reentrantLock = new ReentrantLock();public static void main(String[] args) throws InterruptedException {Condition condition1 = reentrantLock.newCondition();Condition condition2 = reentrantLock.newCondition();LockEx lockEx1 = new LockEx(str1, reentrantLock, condition1, condition2);Thread thread1 = new Thread(lockEx1);LockEx lockEx2 = new LockEx(str2, reentrantLock, condition2, condition1);Thread thread2 = new Thread(lockEx2);thread1.start();thread2.start();thread1.join();thread2.join();}
}class LockEx implements Runnable {private String str;private ReentrantLock lock;private Condition condition;private Condition other;public LockEx(String str, ReentrantLock lock, Condition condition, Condition other) {this.str = str;this.lock = lock;this.condition = condition;this.other = other;}@Overridepublic void run() {int length = str.length();lock.lock();try {for (int i = 0; i < length; i++) {System.out.println("线程" + Thread.currentThread().getId() + ":" + str.charAt(i));other.signalAll();condition.await();}other.signalAll();} catch (Exception e) {e.printStackTrace();} finally {lock.unlock();}}
}

3.数组升序排列

这里排序算法都行，以冒泡为例

public class Bubble {@Testpublic void test() {int nums[] = {1, 2, 3, 4, 5, 6};bubble(nums);System.out.println("最终是：" + Arrays.toString(nums));}public void bubble(int nums[]) {int length = nums.length;boolean change = false;for (int i = 0; i < length - 1; i++) {for (int j = 0; j < length - 1 - i; j++) {if (nums[j] > nums[j + 1]) {swap(nums, j, j + 1);change = true;}}System.out.println("第" + (i + 1) + "次循环后结果是:" + Arrays.toString(nums));if (change == false) {break;} else {change = true;}}}public void swap(int nums[], int x, int y) {int temp = nums[x];nums[x] = nums[y];nums[y] = temp;}
}