• 109th LeetCode Weekly Contest 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:

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

    就是说有一个数组,里面的每一个数字-3000得到的结果,在数组里面>=这个结果的有多少个

    class RecentCounter {
    public:
        int x = 0;
        int num[20000];
        int pre = 0;
        //int y;
        RecentCounter() {
    
        }
    
        int ping(int t) {
            int result = 0;
            num[pre] = t;
            pre ++;
            x = t - 3000;
            for(int i = 0; i<pre; i++){
                if(num[i]>=x){
                    result++;
                }
            }
            return result;
        }
    };
    /**
     * Your RecentCounter object will be instantiated and called as such:
     * RecentCounter* obj = new RecentCounter();
     * int param_1 = obj->ping(t);
     */
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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/9904053.html
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