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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
object Solution {
def maxSlidingWindow(nums: Array[Int], k: Int): Array[Int] = {
val q = new java.util.ArrayDeque[(Int, Int)]()
def populate(num: Int, index: Int): Unit = {
while (!q.isEmpty && q.peekLast()._2 + k <= index) {
q.pollLast()
}
while (!q.isEmpty && (q.peekFirst()._1 <= num)) {
q.pollFirst()
}
q.addFirst((num, index))
}
if (k == 1) {
nums
} else {
for (i <- (0 until k - 1)) {
populate(nums(i), i)
}
((k - 1) until nums.size).map { i =>
populate(nums(i), i)
q.peekLast()._1
}.toArray
}
}
}