239. Sliding Window Maximum

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.

Example 1:

Input: nums = [1,3,-1,-3,5,3,6,7], 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**

Example 2:

Input: nums = [1], k = 1
Output: [1]

Example 3:

Input: nums = [1,-1], k = 1
Output: [1,-1]

Example 4:

Input: nums = [9,11], k = 2
Output: [11]

Example 5:

Input: nums = [4,-2], k = 2
Output: [4]

Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104
• 1 <= k <= nums.length

# @lc code=start
using LeetCode
function max_sliding_window(nums::Vector{Int}, k::Int)::Vector{Int}
    q = Deque{Int}()
    len = length(nums)
    res = fill(0, len - k + 1)
    for i in 1:k-1
        num = nums[i]
        while !isempty(q) && nums[first(q)] < num
            popfirst!(q)
        end
        push!(q, i)
    end
    for i in k:len
        num = nums[i]
        while !isempty(q) && (nums[first(q)] < num || first(q) < i - k + 1)
            popfirst!(q)
        end
        while !isempty(q) && nums[last(q)] < num
            pop!(q)
        end
        push!(q, i)
        res[i - k + 1] = nums[first(q)]
    end
    res
end
# @lc code=end

max_sliding_window (generic function with 1 method)

This page was generated using DemoCards.jl and Literate.jl.