原题
我们提供一个类:
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");
}
}
}
两个不同的线程将会共用一个 FooBar
实例。其中一个线程将会调用 foo()
方法,另一个线程将会调用 bar()
方法。
请设计修改程序,以确保 "foobar" 被输出 n 次。
示例 1:
输入: n = 1
输出: "foobar"
解释: 这里有两个线程被异步启动。其中一个调用 foo() 方法, 另一个调用 bar() 方法,"foobar" 将被输出一次。
示例 2:
输入: n = 2
输出: "foobarfoobar"
解释: "foobar" 将被输出两次。
解法
思想
wait_notify, await_signal
代码
wait_notify
class FooBar {
boolean flag = false;
private int n;
public FooBar(int n) {
this.n = n;
}
public void foo(Runnable printFoo) throws InterruptedException {
synchronized(this){
for (int i = 0; i < n; i++) {
while(flag) this.wait();
// printFoo.run() outputs "foo". Do not change or remove this line.
printFoo.run();
flag = true;
this.notifyAll();
}
}
}
public void bar(Runnable printBar) throws InterruptedException {
synchronized(this){
for (int i = 0; i < n; i++) {
while(!flag) this.wait();
// printBar.run() outputs "bar". Do not change or remove this line.
printBar.run();
flag = false;
this.notifyAll();
}
}
}
}
await_signal
class FooBar {
private int n;
boolean flag;
ReentrantLock lock;
Condition condition;
public FooBar(int n) {
this.n = n;
lock = new ReentrantLock();
condition = lock.newCondition();
}
public void foo(Runnable printFoo) throws InterruptedException {
lock.lock();
try{
for (int i = 0; i < n; i++) {
while(flag) condition.await();
// printFoo.run() outputs "foo". Do not change or remove this line.
printFoo.run();
flag = true;
condition.signalAll();
}
}finally{
lock.unlock();
}
}
public void bar(Runnable printBar) throws InterruptedException {
lock.lock();
try{
for (int i = 0; i < n; i++) {
while(!flag) condition.await();
// printBar.run() outputs "bar". Do not change or remove this line.
printBar.run();
flag = false;
condition.signalAll();
}
}finally{
lock.unlock();
}
}
}
原创文章,作者:彭晨涛,如若转载,请注明出处:https://www.codetool.top/article/leetcode1115-%e4%ba%a4%e6%9b%bf%e6%89%93%e5%8d%b0foobar/