ou have a RecentCounter
class which counts the number of recent requests within a certain time frame.
Implement the RecentCounter
class:
RecentCounter()
Initializes the counter with zero recent requests.int ping(int t)
Adds a new request at time t
, where t
represents some time in milliseconds, and returns the number of requests that has happened in the past 3000
milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range [t - 3000, t]
.It is guaranteed that every call to ping
uses a strictly larger value of t
than the previous call.
Example 1:
Input ["RecentCounter", "ping", "ping", "ping", "ping"] [[], [1], [100], [3001], [3002]] Output [null, 1, 2, 3, 3] Explanation RecentCounter recentCounter = new RecentCounter(); recentCounter.ping(1); // requests = [1], range is [-2999,1], return 1 recentCounter.ping(100); // requests = [1, 100], range is [-2900,100], return 2 recentCounter.ping(3001); // requests = [1, 100, 3001], range is [1,3001], return 3 recentCounter.ping(3002); // requests = [1, 100, 3001, 3002], range is [2,3002], return 3
Constraints:
1 <= t <= 109
ping
with strictly increasing values of t
.104
calls will be made to ping
.这题就是让写一个方法,每次call这个方法,能return之前call这个方法加入的数字里,比当前call的这个数字少3000以内的个数。
直觉就是用一个queue,刚开始还想着用pq,但是因为这个数字是严格递增的,所以大可不必qp了,普通queue就行了,默认排好序了。于是就被卡在了queue的api上,刚开始还想着要怎么遍历queue,后来看了解析才想到直接while (queue.peek())就行,把 < t - 3000的都扔掉,最后return queue size就行。
- class RecentCounter {
- Queue
queue = new LinkedList<>(); -
- public RecentCounter() {
-
- }
-
- public int ping(int t) {
- queue.add(t);
- while (queue.peek() < t - 3000) {
- queue.remove();
- }
- return queue.size();
- }
- }
-
- /**
- * Your RecentCounter object will be instantiated and called as such:
- * RecentCounter obj = new RecentCounter();
- * int param_1 = obj.ping(t);
- */
这个牛逼的解析里有很多不同的方法,可以借鉴一下,但是我懒得看了:LeetCode - The World's Leading Online Programming Learning Platform