239 Sliding Window Maximum¶
Problem:¶
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
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?
Solutions:¶
public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int[0];
}
int[] res = new int[nums.length - k + 1];
PriorityQueue<Integer> q = new PriorityQueue<Integer>(Collections.reverseOrder());
for (int i = 0; i < k; i ++) {
q.add(nums[i]);
}
res[0] = q.peek();
for (int j = k; j < nums.length; j ++) {
int remove = nums[j - k];
q.add(nums[j]);
q.remove((Integer)remove);
res[j - k + 1] = q.peek();
}
return res;
}
}
public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int[0];
}
int[] res = new int[nums.length - k + 1];
LinkedList<Integer> q = new LinkedList<Integer>();
for (int i = 0; i < nums.length; i ++) {
if (!q.isEmpty() && q.peek() == i - k) {
q.removeFirst();
}
while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
q.removeLast();
}
q.add(i);
if (i + 1 >= k) {
res[i + 1 - k] = nums[q.peekFirst()];
}
}
return res;
}
}