239. Sliding Window Maximum

Question

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

Thinking:

Method 1:优先级队列。

class Solution {
  public int[] maxSlidingWindow(int[] nums, int k) {
      if(nums == null || nums.length == 0) return nums;
      PriorityQueue<Integer> pq = new PriorityQueue<>(new Comparator<Integer>(){
          @Override
          public int compare(Integer n1, Integer n2){
              return n2 - n1;
          }
      });
      for(int i = 0; i < k; i++)
          pq.offer(nums[i]);
      int[] result = new int[nums.length - k + 1];
      for(int i = k; i <= nums.length; i++){
          result[i - k] = pq.peek();
          pq.remove(nums[i - k]);
          if(i < nums.length) pq.offer(nums[i]);
      }
      return result;
  }
}

二刷

Still use priority queue. It’s not O(n);

class Solution {
 public int[] maxSlidingWindow(int[] nums, int k) {
     if(nums == null || nums.length == 0) return new int[0];
     int len = nums.length;
     int[] result = new int[len - k + 1];
     PriorityQueue<Integer> pq = new PriorityQueue<>(new Comparator<Integer>(){
         @Override
         public int compare(Integer a, Integer b){
             return b - a;
         }
     });
     for(int i = 0; i < k; i++){
         pq.offer(nums[i]);
     }
     for(int i = k; i <= nums.length; i++){
         result[i - k] = pq.peek();
         pq.remove(nums[i - k]);
         if(i < nums.length)
             pq.offer(nums[i]);
     }
     return result;
 }
}

Amazon session

Method 1: Deque

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int len = nums.length;
        if(k == 0) return new int[0];
        int[] res = new int[len - k + 1];
        Deque<Integer> q = new LinkedList<>();  // q saves the index
        for(int i = 0; i < len; i++){
            if(!q.isEmpty() && q.peek() == i - k) q.poll();
            while(!q.isEmpty() && nums[i] >= nums[q.peekLast()]) q.pollLast();
            q.offer(i);
            if(i - k + 1 >= 0) res[i - k + 1] = nums[q.peek()];
        }
        return res;
    }
}