LWC 109: 933.Number of Recent Calls

Posted 2023-02-07 Demon的黑与白

LWC 109: 933.Number of Recent Calls

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Problem:

Write a class RecentCounter to count recent requests. It has only one method: ping(int t), where t represents some time in milliseconds. Return the number of pings that have been made from 3000 milliseconds ago until now. Any ping with time in [t - 3000, t] will count, including the current ping. It is guaranteed that every call to ping uses a strictly larger value of t than before.

Example 1:

Input: K = 0
Output: 5
Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.

Example 2:

Input: inputs = [“RecentCounter”,“ping”,“ping”,“ping”,“ping”], inputs = [[],1,[100],[3001],[3002]]
Output: [null,1,2,3,3]

Note:

• Each test case will have at most 10000 calls to ping.
• Each test case will call ping with strictly increasing values of t.
• Each call to ping will have 1 <= t <= 10^9.

思路：
暴力解法，先把一个个时间戳记录下来，每当新来一个时间戳时统计[t - 3000, t]范围内的个数。

Version 1：

public class RecentCounter 
	
	List<Integer> ans;
	public RecentCounter() 
        ans = new ArrayList<>();
    
    
    public int ping(int t) 
    	ans.add(t);
    	int cnt = 0;
    	int s = t - 3000;
    	int e = t;
    	for (int v : ans) 
    		if (v >= s && v <= e) cnt ++;
    	
    	return cnt;
    

优化一下，传入新的时间戳后可以保持记录数组的有序性，采用二分快速查找find the last value x in array where x + 3000 < t

Version2：

public class RecentCounter 
	
	int[] mem;
	int n;
	public RecentCounter() 
		mem = new int[10000 + 12];
		n = 0;
	
    
    public int ping(int t) 
    	mem[n++] = t;
    	int c = n - bs(mem, t, 0, n) - 1;
    	return c;
    
    
    public int bs(int[] mem, int t, int s, int e) 
    	int lf = s;
    	int rt = e - 1;
    	// x + 3000 < t
    	while (lf < rt) 
    		int mid = lf + (rt - lf + 1) / 2;
    		int val = mem[mid] + 3000;
    		if (val < t) 
    			lf = mid;
    		
    		else  // >= t
    			rt = mid - 1;
    		
    	
    	if (mem[lf] + 3000 >= t) return -1;
    	return lf;
    

当然我们还可以维护一个单调递增的队列，并实时过滤头部元素~

Version3：

import java.util.ArrayDeque;
import java.util.Deque;

public class RecentCounter 
	
	Deque<Integer> pq;
	public RecentCounter() 
		pq = new ArrayDeque<>();
	
    
    public int ping(int t) 
    	while (!pq.isEmpty() && pq.peekFirst() + 3000 < t) 
    		pq.pollFirst();
    	
    	pq.addLast(t);
    	return pq.size();