题目地址：Print FooBar Alternately - LeetCode
Suppose you are given the following code:
class FooBar {
public void foo() {
for (int i = 0; i < n; i++) {
print("foo");
}
}
public void bar() {
for (int i = 0; i < n; i++) {
print("bar");
}
}
}
The same instance of FooBar will be passed to two different threads. Thread A will call foo() while thread B will call bar(). Modify the given program to output “foobar” n times.
Example 1:
Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar(). "foobar" is being output 1 time.
Example 2:
Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.
这道题目的意思是多线程情况下如何保证代码执行的顺序，是一道经典的多线程的题目。
最容易想到的是用Java的synchronized加锁，然后flag变量控制执行顺序。
Java解法如下：
class FooBar {
volatile private int n;
private int flag = 0;
public FooBar(int n) {
this.n = n;
}
public void foo(Runnable printFoo) throws InterruptedException {
for (int i = 0; i < n; i++) {
synchronized (this) {
while (flag == 1) {
this.wait();
}
// printFoo.run() outputs "foo". Do not change or remove this line.
printFoo.run();
flag = 1;
this.notifyAll();
}
}
}
public void bar(Runnable printBar) throws InterruptedException {
for (int i = 0; i < n; i++) {
synchronized (this) {
while (flag == 0) {
this.wait();
}
// printBar.run() outputs "bar". Do not change or remove this line.
printBar.run();
flag = 0;
this.notifyAll();
}
}
}
}
当然我们可以使用更高级的工具，比如说 CyclicBarrier。
注意：下面的代码有问题，不能确保并发的顺序！！
class FooBar {
private int n;
private final CyclicBarrier barrier = new CyclicBarrier(2);
public FooBar(int n) {
this.n = n;
}
public void foo(Runnable printFoo) throws InterruptedException {
for (int i = 0; i < n; i++) {
// printFoo.run() outputs "foo". Do not change or remove this line.
printFoo.run();
try {
barrier.await();
} catch (BrokenBarrierException e) {
e.printStackTrace();
}
}
}
public void bar(Runnable printBar) throws InterruptedException {
for (int i = 0; i < n; i++) {
try {
barrier.await();
} catch (BrokenBarrierException e) {
e.printStackTrace();
}
// printBar.run() outputs "bar". Do not change or remove this line.
printBar.run();
}
}
}
另一种正确的做法是使用信号量：Semaphore
class FooBar {
private int n;
private final Semaphore foo = new Semaphore(0);
private final Semaphore bar = new Semaphore(1);
public FooBar(int n) {
this.n = n;
}
public void foo(Runnable printFoo) throws InterruptedException {
for (int i = 0; i < n; i++) {
bar.acquire();
// printFoo.run() outputs "foo". Do not change or remove this line.
printFoo.run();
foo.release();
}
}
public void bar(Runnable printBar) throws InterruptedException {
for (int i = 0; i < n; i++) {
foo.acquire();
// printBar.run() outputs "bar". Do not change or remove this line.
printBar.run();
bar.release();
}
}
}
文档信息
- 本文作者：last2win
- 本文链接：https://last2win.com/2020/01/12/LeetCode/
- 版权声明：自由转载-非商用-非衍生-保持署名（创意共享3.0许可证）