239. Sliding Window Maximum (1)

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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?

Solution

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        
        int len = nums.length;
        if(len == 0)
            return nums;
        
        int[] result = new int[len-k+1];
        LinkedList<Integer> list = new LinkedList<>();//for all promising index of max value
        
        for(int i = 0; i < len; i ++)
        {
            //Skip indexes who is out of range
            while(!list.isEmpty() && list.peek() < i-k+1){
                list.poll();
            }
            //Skip all index which is smaller than current value
            while(!list.isEmpty()&& nums[list.peekLast()] < nums[i]){
                list.pollLast();
            }
            list.add(i);
            
            //While reach to each boundary (the last index of window
            //shall update the result
            if(i+1 >= k){
                result[i+1-k] = nums[list.peek()];
            }

        }
        return result;
    }
}