leetcode 933.Number of Recent Calls 简单
Number of Recent Calls
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: 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
题目叙述的醉了。
会给你系列的Ping的时间,以毫秒为单位,然后保证递增,ping函数返回当前的输入的时间t的3000毫秒以内的Ping的个数。
我犯了错,应该是当队列不为空的时候循环,而不是for(int i =0;i<q.size();++i)
class RecentCounter {
private:
queue<int> q;
public:
RecentCounter() {
}
int ping(int t) {
while(!q.empty()){
if( t-q.front()> 3000){
q.pop();
} else {
break;
}
}
q.push(t);
return q.size();
}
};
/**
* Your RecentCounter object will be instantiated and called as such:
* RecentCounter* obj = new RecentCounter();
* int param_1 = obj->ping(t);
*/