Sliding Window Maximum
Given an array 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.
For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
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
Therefore, return the max sliding window as [3,3,5,5,6,7].
Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.
Follow up:
Could you solve it in linear time?
public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int[0];
}
Deque<Integer> deque = new LinkedList<>();
int[] result = new int[nums.length - k + 1];
for (int i = 0; i < nums.length; i++) {
if (!deque.isEmpty() && deque.peek() == i - k) {
deque.poll();
}
while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
deque.pollLast();
}
deque.offer(i);
if (i >= k - 1) {
result[i - k + 1] = nums[deque.peek()];
}
}
return result;
}
}
Hope this helps,
Michael
