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import java.util.*;

/**

* TestCases

[1,-1] 1

[7,2,4] 2

[1,3,1,2,0,5] 3

*/

public class SlidingWindowMaximum {

/**

* 注意PriorityQueue的remove复杂度是O(k)，所以本题复杂度是O(n*k)

* 可以将PriorityQueue转成TreeMap，复杂度为O(n*lgk)

*/

public int[] maxSlidingWindow(int[] nums, int k) {

if (nums.length == 0 || k == 0) {

return new int[0];

}

int[] result = new int[nums.length - k + 1];

Queue<Integer> queue = new PriorityQueue<>(Comparator.<Integer>reverseOrder());

for (int i = 0; i < nums.length; i++) {

queue.offer(nums[i]);

if (queue.size() > k) {

queue.remove(nums[i - k]);

}

if (i - k + 1 >= 0) {

result[i - k + 1] = queue.peek();

}

}

return result;

}

public int[] maxSlidingWindow2(int[] nums, int k) {

if (nums.length == 0 || k == 0) {

return new int[0];

}

int[] result = new int[nums.length - k + 1];

Deque<Integer> queue = new LinkedList<>();

for (int i = 0; i < nums.length; i++) {

while (!queue.isEmpty() && nums[i] > nums[queue.peekLast()]) {

queue.pollLast();

}

queue.offerLast(i);

if (queue.peekFirst() <= i - k) {

queue.pollFirst();

}

if (i - k + 1 >= 0) {

result[i - k + 1] = nums[queue.peekFirst()];

}

}

return result;

}

}