Question

link

有一个接口叫 void setRPS(int num);

接下来不断有 request 过来，如何实现下面的接口，返回 accept 或者 deny，

bool process(int timestamp){

}

Solution

Suggested by level 5 of this post:

  1. maintain a variable for the number of request processed/rejected.
    1. This variable must be atomic, thus a AtomicInteger.
    2. the variable is called ‘count’
  2. have a method to process request
    1. if count < limit, do it
    2. otherwise, reject
  3. This is the most important: clear the count every 1 seconds!
    1. eg. LIMIT = 5r/s, so:
    2. the first 5 number of requests in every second are getting fulfilled
    3. from 6th request onward, the request all rejected, until the next second.

Code

public class SetRps {

    AtomicInteger count = new AtomicInteger(0);
    int limit = -1;
    int printIndex = 1;
    long startTimestamp = -1;

    void setRPS(int num) {
        limit = num;
    }

    boolean process(long timestamp) {
        // suppose timestamp is ms
        synchronized (this) {
            // to process or not to process
            if (count.get() < limit) {
                // can process
                count.incrementAndGet();
                System.out.println(printIndex++ + ". processing request "
                        + timestamp % 100000 / 1000 + "," + timestamp % 1000);
                return true;
            }
            // to clear or not to clear
            if (timestamp - startTimestamp >= 1000) {
                // every 1 seconds, reset
                count.set(0);
                startTimestamp = timestamp;
                System.out.println("clear!");
                return true;
            }
        }
        return false;
    }
}