239. Sliding Window Maximum (Hard)
https://leetcode.com/problems/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?
Solutions
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int[0];
}
int len = nums.length;
int[] ans = new int[len - k + 1];
int index = 0;
// Use a deque to store index
Deque<Integer> qmax = new ArrayDeque<>();
for (int i = 0; i < len; i++) {
// Remove the head of deque when index equals to (i-k) since it's out of window k
if (!qmax.isEmpty() && qmax.getFirst() == i - k) {
qmax.removeFirst();
}
// The following statement will not increase the time complexity since the maximum times of
// operations of adding elements to qmax is less than length of nums. Therefor, the time
// complexity of this solution is at most O(2n)=O(n)
// Remove numbers that no bigger than nums[i] in window k as they are negligible and useless.
while (!qmax.isEmpty() && nums[qmax.getLast()] <= nums[i]) {
qmax.removeLast();
}
// The head of qmax must be >= nums[i] since it's not removed. Index in the front of qmax
// must be local maximum except these before it
qmax.addLast(i);
// The deque qmax keeps index while the array ans keeps value
if (i >= k - 1) {
ans[index++] = nums[qmax.getFirst()];
}
}
return ans;
}
}