Update _239.java

Anacoder1 · web-flow · commit ebda5b312b1b · 2020-06-21T23:20:46.000+05:30
diff --git a/src/main/java/com/fishercoder/solutions/_239.java b/src/main/java/com/fishercoder/solutions/_239.java
@@ -1,60 +1,28 @@
-package com.fishercoder.solutions;
-
-import java.util.PriorityQueue;
-
-/**
- * 239. 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?
-
- 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.
- */
-public class _239 {
-
-    public static class Solution1 {
-        public int[] maxSlidingWindow(int[] nums, int k) {
-            if (nums == null || nums.length == 0 || k == 0) {
-                return new int[0];
-            }
-            PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
-            int[] res = new int[nums.length - k + 1];
-            for (int i = 0; i < nums.length; i++) {
-                if (i < k) {
-                    heap.offer(nums[i]);
-                    if (i == k - 1) {
-                        res[0] = heap.peek();
-                    }
-                } else {
-                    heap.remove(nums[i - k]);
-                    heap.offer(nums[i]);
-                    res[i - k + 1] = heap.peek();
-                }
-            }
-            return res;
-        }
-    }
+public class Solution {
+    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;
+	}
 }
