启动N个线程, 这N个线程要不间断按顺序打印数字1-N. 将问题简化为3个线程无限循环打印1到3
方法一: 使用synchronized
三个线程无序竞争同步锁, 如果遇上的是自己的数字, 就打印. 这种方式会浪费大量的循环
public class TestSequential1 {
private volatile int pos = 1;
private volatile int count = 0;
public void one(int i) {
synchronized (this) {
if (pos == i) {
System.out.println("T-" + i + " " + count);
pos = i % 3 + 1;
count = 0;
} else {
count++;
public static void main(String[] args) {
TestSequential1 demo = new TestSequential1();
for (int i = 1; i <=3; i++) {
int j = i;
new Thread(()->{
while(true) {
demo.one(j);
}).start();
T-1 0
T-2 5793
T-3 5285
T-1 2616
T-2 33
T-3 28
T-1 22
T-2 44
T-3 6
T-1 881
T-2 118358
T-3 247380
T-1 30803
T-2 29627
T-3 52044
方法二: 使用synchronized配合wait()和notifyAll()
竞争同步锁时使用wait()和notifyAll(), 可以避免浪费循环
public class TestSequential01 {
private volatile int pos = 1;
private volatile int count = 0;
private final Object obj = new Object();
public void run(int i) {
int next = i % 3 + 1;
while(true) {
synchronized (obj) {
System.out.println(i + " in");
try {
while (pos != i) {
count++;
System.out.println(i + " wait");
obj.wait();
System.out.println("T-" + i + " " + count);
pos = next;
count = 0;
obj.notifyAll();
} catch (InterruptedException e) {
e.printStackTrace();
public static void main(String[] args) {
TestSequential01 demo = new TestSequential01();
for (int i = 3; i >=1; i--) {
int j = i;
new Thread(()->{
demo.run(j);
}).start();
3 wait
T-1 1
1 wait
T-2 1
2 wait
1 wait
T-3 2
3 wait
T-1 1
1 wait
T-2 1
2 wait
1 wait
T-3 2
3 wait
T-1 1
1 wait
方法三: 使用可重入锁
用Lock做, 非公平锁, 三个线程竞争, 如果遇上的是自己的数字, 就打印. 这种方式会浪费大量的循环
public class TestSequential01 {
private final Lock lock = new ReentrantLock();
private volatile int pos = 1;
private volatile int count = 0;
public void run(int i) {
int next = i % 3 + 1;
while(true) {
lock.lock();
if (pos == i) {
System.out.println("T-" + i + " " + count);
pos = next;
count = 0;
} else {
count++;
lock.unlock();
public static void main(String[] args) {
TestSequential01 demo = new TestSequential01();
for (int i = 1; i <=3; i++) {
int j = i;
new Thread(()->{
demo.run(j);
}).start();
T-1 0
T-2 0
T-3 323
T-1 54
T-2 68964
T-3 97642
T-1 6504
T-2 100603
T-3 6989
T-1 1313
T-2 0
T-3 183741
T-1 233
T-2 5081
T-3 164367
方法四: 使用可重入锁, 启用公平锁
和3一样, 但是使用公平锁, 这种情况下基本上可以做到顺序执行, 偶尔会产生多一次循环
private final Lock lock = new ReentrantLock(true);
T-1 0
T-2 0
T-3 0
T-1 0
T-2 0
T-3 0
T-1 0
T-2 0
T-3 0
T-1 0
T-2 0
T-3 1
T-1 1
T-2 1
T-3 1
方法五: 使用Condition
给每个线程不同的condition. 可以用condition.signal()精确地通知对应的线程继续执行(在对应的condition上await的线程, 可能是多个).
public class TestSequential01 {
private static Lock lock = new ReentrantLock();
private static Condition[] conditions = {lock.newCondition(), lock.newCondition(), lock.newCondition()};
private volatile int state = 1;
private void run(final int self) {
int next = self % 3 + 1;
while(true) {
lock.lock();
try {
while(this.state != self) {
conditions[self - 1].await();
System.out.println(self);
this.state = next;
conditions[next - 1].signal();
} catch (InterruptedException e) {
e.printStackTrace();
lock.unlock();
public static void main(String[] args) {
TestSequential01 rlc = new TestSequential01();
for (int i = 1; i < 4; i++) {
int j = i;
new Thread(()->rlc.run(j)).start();
在使用wait()和await()的竞争环境, 因为被notifyAll()和signal()之后到线程回到执行之前, 条件可能发生变化, 所以必须在wait()和await()外包使用while循环检测条件, 这是一个通用方法
