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. Return the max sliding window.

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

Note: 
You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.

Follow up:
Could you solve it in linear time?

 

Approach #1: C++. [brute force]

class Solution {
public:
    vector<int> maxSlidingWindow2(vector<int>& nums, int k) {
        vector<int> ans;
        if (nums.empty()) return ans;
        for (int i = 0; i + k <= nums.size(); ++i) {
            vector<int> temp(nums.begin()+i, nums.begin()+i+k);
            ans.push_back(*max_element(temp.begin(), temp.end()));
        }
        return ans;
    }
};

　　

Approach #1: Java. [deque]

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || k < 0) return new int[0];
        
        int n = nums.length;
        int[] r = new int[n-k+1];
        int ri = 0;
        
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            while (!q.isEmpty() && q.peek() < i-k+1) 
                q.poll();
            while (!q.isEmpty() && nums[q.peekLast()] < nums[i])
                q.pollLast();
            q.offer(i);
            if (i >= k-1) {
                r[ri++] = nums[q.peek()];
            }
        }
        return r;
    }
}

　　

Appraoch #3: Python. [deque]

from collections import deque
class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        if not nums:
            return []
        if k == 0:
            return nums
        deq = deque()
        result = []
        
        for i in range(len(nums)):
            while len(deq) != 0 and deq[0] < i-k+1:
                deq.popleft()
            while len(deq) != 0 and nums[i] > nums[deq[-1]]:
                deq.pop()
            deq.append(i)
            
            if i >= k-1:
                result.append(nums[deq[0]])
        
        return result