933. Number of Recent Calls

#### QUESTION:

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 `ping`s 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`.

#### EXPLANATION:

1.先将这次的t放在最后

2.接着往前遍历到t-3000的地方

3.计算sum

#### SOLUTION:

``````//我的解决方案
class RecentCounter {

ArrayDeque<Integer> queue;

public RecentCounter() {
queue = new ArrayDeque<>();
}

public int ping(int t) {
int sum = 0;
Iterator<Integer> iterator = queue.descendingIterator();
while (iterator.hasNext() && iterator.next() >= t-3000){
sum++;
}
return sum;
}

}

//AC中最快的解
class RecentCounter {
private static final int N = 10000;
private final int array[] = new int[N];
private int idx = 0;
private int lowIdx = -1;

public RecentCounter() {
}

public int ping(int t) {
int ans = 0;
array[idx] = t;
if (lowIdx < 0) {
lowIdx = idx;
ans = 1;
} else {
while(array[idx] - array[lowIdx] > 3000) {
lowIdx++;
}
ans = (idx - lowIdx + 1);
}
idx++;
return ans;
}
}
``````
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