Leetcode 239 Sliding Window Maximum (designated sliding window maximum value) (sliding window)

Leetcode

Problem Description

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. Return the max sliding window.

example

Example:
Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 
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

Problem-solving approach

  First generation deque deque, deque index for storing an array of nums.

1. If the deque is empty, directly to the subscript i deque discharge, into the end of the process;
2. If the deque is not empty, remove the tail current deque stored in the table, is assumed to be J;
   . 1) if nums [j]> nums [i ], the subscript i directly into the tail, into the end of the process;.
   2 .) If nums [j] <= nums [ i], j ejected from the deque is continued into the deque rules;
3. If the index is equal to the first team deque iw, describes the current index deque team head has expired, eject the current subscript team head.

** Solution Java (deque) **
** 8ms, beats 87.35% **
**49.5, beats 6.25% **
public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length < k || k < 1)
            return new int[0];
        Deque<Integer> deque = new ArrayDeque<>();
        int index = 0;
        int[] ans = new int[nums.length + 1 - k];
        for (int i = 0; i < nums.length; ++i) {
            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i])
                deque.pollLast();
            deque.offer(i);
            if (deque.peek() == i - k)
                deque.poll();
            if (i >= k - 1)
                ans[index++] = nums[deque.peek()];
        }
        return ans;
    }
}

** Solution Java LinkedList **
** 6ms, beats 90.50% **
** 47.8MB, beats 6.25% **
class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || k<1 || nums.length < k)
            return new int[0];
        LinkedList<Integer> list = new LinkedList<>();
        int[] ans = new int[nums.length - k + 1];
        int index = 0;
        for (int i = 0; i < nums.length; ++i) {
            while (!list.isEmpty() && nums[list.peekLast()] < nums[i])   
                list.pollLast();
            list.addLast(i);
            if (list.peekFirst() == i - k)
                list.pollFirst();
            if (i >= k - 1)
                ans[index++] = nums[list.peekFirst()];
        }
        return ans;
    }
}

** Solution Python3 **
** 176ms, 49.00% **
** 19.5MB, 88.46% **
class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        if (nums == None or len(nums) < k or k < 1) :
            return []
        ans = []
        q = collections.deque()
        for i in range(len(nums)) :
            while (len(q) != 0 and nums[q[-1]] < nums[i]) :
                q.pop()
            q.append(i)
            if (q[0] == i - k) :
                q.popLeft()
            if (i >= k - 1) :
                ans.append(nums[q[0]])
        return ans;