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);
 */