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