需求描述
写一个1秒两个的限流工具类，2r/s
使用semaphore
代码实现-类似令牌桶算法
public class LimitHelper {private int maxLimit;private Semaphore semaphore;private int timeoutSeconds;public LimitHelper(int maxLimit, int timeoutSeconds) {this.maxLimit = maxLimit;semaphore = new Semaphore(maxLimit);this.timeoutSeconds = timeoutSeconds;this.autoRelease();}/*** 每秒钟释放两个信号出来*/private void autoRelease() {new Thread(() -> {try {while (true) {TimeUnit.SECONDS.sleep(timeoutSeconds);if (!semaphore.tryAcquire(1)) {// 无信号了semaphore.release(maxLimit);}else {// 池中有信号，并且消耗了一个，释放一个补偿semaphore.release(1);}}} catch (InterruptedException e) {e.printStackTrace();}}).start();}public boolean acquire() {try {return semaphore.tryAcquire(1, timeoutSeconds, TimeUnit.SECONDS);} catch (InterruptedException e) {throw new RuntimeException(e);}}public static void main(String[] args) {LimitHelper limitHelper = new LimitHelper(2, 1);for (int i = 0; i < 5; i++) {new Thread(() -> {while (true) {if (limitHelper.acquire()) {System.out.println( System.currentTimeMillis() / 1000+ " " + Thread.currentThread().getName() + " 获取到令牌");try {// 业务处理TimeUnit.MILLISECONDS.sleep(20);} catch (InterruptedException e) {throw new RuntimeException(e);}} else {System.out.println(System.currentTimeMillis() / 1000 + " "+Thread.currentThread().getName() + " 被限流了");}}}).start();}}}结果演示
滑动窗口算法实现
如上图，红色是被限流的请求事件，蓝色是允许的请求事件。
代码实现
public class LimitHelper2 {/*** 次数*/private int maxRate;/*** 多久时间内*/private int rangeTimeSeconds;/*** 历史记录*/private volatile LinkedList<Long> timeRecord;public LimitHelper2(int maxRate, int rangeTimeSeconds) {this.maxRate = maxRate;this.rangeTimeSeconds = rangeTimeSeconds;timeRecord = new LinkedList<>();}/*** 2r/s 限流实现*/private synchronized boolean acquire() {long now = System.currentTimeMillis();long min = now - (rangeTimeSeconds * 1000L);if (timeRecord.isEmpty()) {timeRecord.addLast(now);return true;}// 不为空并且size >= maxRateint count = 0;if (timeRecord.size() >= maxRate) {LinkedList<Long> newRecords = new LinkedList<>();for (int i = 0; i < maxRate; i++) {Long beforeHappenTime = timeRecord.pollLast();if (beforeHappenTime == null) {timeRecord.addLast(now);System.out.println("before时间为null");return true;}newRecords.addFirst(beforeHappenTime);if (beforeHappenTime.compareTo(min) >= 0) {count++;} else {break;}}timeRecord = newRecords;if (count >= maxRate) {return false;} else {timeRecord.addLast(now);return true;}} else {timeRecord.add(now);return true;}}public static void main(String[] args) {// 1秒2次 限流LimitHelper2 limiter = new LimitHelper2(4, 1);for (int i = 0; i < 5; i++) {new Thread(() -> {while (true) {if (limiter.acquire()) {try {System.out.println(System.currentTimeMillis() + " " + Thread.currentThread().getName() + " 执行任务");Thread.sleep(10);} catch (InterruptedException e) {e.printStackTrace();}} else {System.out.println(System.currentTimeMillis() + " " + Thread.currentThread().getName() + " 被限流");}try {Thread.sleep(500);} catch (InterruptedException e) {throw new RuntimeException(e);}}}).start();}}}结果演示
4r/s
