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?
Hint:
How about using a data structure such as deque (double-ended queue)?
The queue size need not be the same as the window’s size.
Remove redundant elements and the queue should store only elements that need to be considered.
Solution 1:
public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if(nums.length==0) return new int[0];
int[] ans = new int[nums.length-k+1];
int left = 0, right = 0;
PriorityQueue<Integer> pq = new PriorityQueue<>(k, (a,b)->(b-a));
while(right<k) pq.offer(nums[right++]);
while(right<=nums.length){
ans[left] = pq.peek();
pq.remove(nums[left++]);
if(right!=nums.length) pq.offer(nums[right]);
right++;
}
return ans;
}
}
Solution 2:
public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if(nums.length==0) return new int[0];
int[] ans = new int[nums.length-k+1];
Deque<Integer> dq = new ArrayDeque<>();
for(int i=0; i<nums.length; i++){
while(!dq.isEmpty() && dq.peek()<i-k+1) dq.poll();
while(!dq.isEmpty() && nums[dq.peekLast()]<=nums[i]) dq.pollLast();
dq.offer(i);
if(i>=k-1) ans[i-k+1] = nums[dq.peek()];
}
return ans;
}
}