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?
public int[] maxSlidingWindow(int[] a, int k) {
int n = a.length;
if(n == 0) return new int[0];
int[] r = new int[n-k+1];
Deque<Integer> q = new ArrayDeque<>();
int ri = 0;
for(int i = 0; i < n; i++){
while(!q.isEmpty() && q.peekFirst() < i - k + 1){ //去除索引过小而超出范围的
q.pollFirst();
}
while (!q.isEmpty() && a[q.peekLast()] < a[i]) { //去除索引比当前新索引小而其值也小的
q.pollLast();
}
// q contains index... r contains content
q.offerLast(i);
if (i >= k - 1) {
r[ri++] = a[q.peek()];
}
}
return r;
}