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.
public int[] maxSlidingWindow(int[] nums, int k) {
if(nums==null || nums.length ==0 || k==0) {
return new int[0];
}
Deque<Integer> window = new LinkedList<Integer>();
int[] result = new int[nums.length > k ? nums.length - k + 1 : 1];
int h = nums.length > k ? k : nums.length;
int i = 0;
for(int j=0;j<nums.length;j++){
while(!window.isEmpty() && nums[window.getLast()]<nums[j]) {
window.removeLast();
}
window.addLast(j);
if(j-window.getFirst()>=k){
window.removeFirst();
}
if(j>=h-1) {
result[i++] = nums[window.getFirst()];
}
}
return result;
}