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Sliding Window Maximum

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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, ie an array with the max of the k numbers within the window, for each position as the window moves across the input. ex: [6,2,1,4] with k=2 would return [6,2,4]

public int[] maxSlidingWindow(int[] a, int k) {		
		if (a == null || k <= 0) {
			return new int[0];
		}
		int n = a.length;
		int[] r = new int[n-k+1];
		int ri = 0;
		// store index
		Deque<Integer> q = new ArrayDeque<>();
		for (int i = 0; i < a.length; i++) {
			// remove numbers out of range k
			while (!q.isEmpty() && q.peek() < i - k + 1) {
				q.poll();
			}
			// remove smaller numbers in k range as they are useless
			while (!q.isEmpty() && a[q.peekLast()] < a[i]) {
				q.pollLast();
			}
			// q contains index... r contains content
			q.offer(i);
			if (i >= k - 1) {
				r[ri++] = a[q.peek()];
			}
		}
		return r;
	}

In the video a linked list is used instead of a dequeue both the concept is the same

Posted by Jamie Meyer 7 months ago

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