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JAVA_59_2.java
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package chapter6;
import java.util.ArrayDeque;
import java.util.Deque;
/**
* 队列的最大值
* <p>
* 请定义一个队列并实现函数max得到队列里的最大值,要求函数max,push_back和pop_front的时间复杂度都是O(1)
* <p>
* 考察点:滑动窗口
*/
public class JAVA_59_2 {
public static void main(String[] args) {
QueueWithMax queueWithMax = new QueueWithMax();
queueWithMax.push_back(2);
System.out.println(queueWithMax.max());
queueWithMax.push_back(3);
System.out.println(queueWithMax.max());
queueWithMax.push_back(4);
System.out.println(queueWithMax.max());
queueWithMax.push_back(2);
System.out.println(queueWithMax.max());
queueWithMax.push_back(6);
System.out.println(queueWithMax.max());
queueWithMax.push_back(2);
System.out.println(queueWithMax.max());
queueWithMax.push_back(5);
System.out.println(queueWithMax.max());
queueWithMax.push_back(1);
System.out.println(queueWithMax.max());
System.out.println("------------");
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
System.out.println("pop " + queueWithMax.pop_front());
System.out.println(queueWithMax.max());
}
public static class QueueWithMax {
private Deque<Integer> data;
private Deque<Integer> maxs;
private int currentIndex = 0;
public QueueWithMax() {
data = new ArrayDeque<>();
maxs = new ArrayDeque<>();
}
public void push_back(Integer element) {
while (!maxs.isEmpty() && element > maxs.peekLast()) {
maxs.pollLast();
}
data.add(element);
maxs.add(element);
currentIndex++;
}
public Integer pop_front() {
if (data.isEmpty()) {
return null;
}
if (maxs.peekFirst().equals(data.peekFirst())) {
maxs.pollFirst();
}
return data.pollFirst();
}
public Integer max() {
if (maxs.isEmpty()) {
return null;
}
return maxs.peekFirst();
}
}
}