LeetCode-in-All
239. Sliding Window Maximum
Hard
You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
--------------- -----
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Example 2:
Input: nums = [1], k = 1
Output: [1]
Example 3:
Input: nums = [1,-1], k = 1
Output: [1,-1]
Example 4:
Input: nums = [9,11], k = 2
Output: [11]
Example 5:
Input: nums = [4,-2], k = 2
Output: [4]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 1041 <= k <= nums.length
Solution
class Solution {
/**
* @param Integer[] $nums
* @param Integer $k
* @return Integer[]
*/
public function maxSlidingWindow($nums, $k) {
$n = count($nums);
$res = array_fill(0, $n - $k + 1, 0);
$x = 0;
$dq = new \SplDoublyLinkedList();
$i = 0;
$j = 0;
while ($j < count($nums)) {
while (!$dq->isEmpty() && $dq->top() < $nums[$j]) {
$dq->pop();
}
$dq->push($nums[$j]);
if ($j - $i + 1 == $k) {
$res[$x] = $dq->bottom();
++$x;
if ($dq->bottom() == $nums[$i]) {
$dq->shift();
}
++$i;
}
++$j;
}
return $res;
}
}