239. Sliding Window Maximum

Given an array of n integer with duplicate number, and a moving window(size k), move the window at each iteration from the start of the array, find the maximum number inside the window at each moving.

Example

For array [1, 2, 7, 7, 8], moving window size k = 3. return [7, 7, 8]

At first the window is at the start of the array like this

[|1, 2, 7| ,7, 8] , return the maximum 7;

then the window move one step forward.

[1, |2, 7 ,7|, 8], return the maximum 7;

then the window move one step forward again.

[1, 2, |7, 7, 8|], return the maximum 8;

Challenge 

o(n) time and O(k) memory

分析：

用LinkedList来保存window里的数，如果新加入的数比LinkedList的最后一个数大，我们可以把LinkedList里比那个新数小的都remove掉。

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
      if (nums == null || k <= 0 || nums.length - k < 0) return new int[0];

      ArrayList<Integer> res = new ArrayList<>();
      // auxiliary queue that is sorted in descending order
      Deque<Integer> q = new LinkedList<>();
      for (int i = 0; i < nums.length; i++) {
          int data = nums[i];
          // enqueue. Remove those smaller values
          while (!q.isEmpty() && q.getLast() < data) {
              q.removeLast();
          }
          q.add(data);
          if (i < k - 1) continue;
          // dequeue. If the current number is the maximum. Also remove it
          // from the queue
          res.add(q.getFirst());
          if (nums[i - k + 1] == q.getFirst()) {
              q.removeFirst();
          }
      }
      return res.stream().mapToInt(i -> i).toArray();
  }
}