239. Sliding Window Maximum

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# 題目敘述

You are given an array of integers 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 1

Input: nums = [1,3,-1,-3,5,3,6,7], 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

# Example 2

Input: nums = [1], k = 1
Output: [1]

# 解題思路

1. Use Deque name dq .
2. Map k integer in the nums , and keep the high number in the dq first value.
3. Adding dp first value in List name res which will become the answer.
4. Making List<Integer> into int[] and return.

# Solution

import java.util.ArrayDeque;

import java.util.ArrayList;

import java.util.Deque;

import java.util.List;

class Solution {

    public int[] maxSlidingWindow(int[] nums, int k) {

        Deque<Integer> dq = new ArrayDeque<>();

        List<Integer> res = new ArrayList<>();

        for (int i = 0; i < k; i++) {

            while (!dq.isEmpty() && nums[i] >= nums[dq.peekLast()]) {

                dq.pollLast();

            }

            dq.offerLast(i);

        }

        res.add(nums[dq.peekFirst()]);

        for (int i = k; i < nums.length; i++) {

            if (dq.peekFirst() == i - k) {

                dq.pollFirst();

            }

            while (!dq.isEmpty() && nums[i] >= nums[dq.peekLast()]) {

                dq.pollLast();

            }

            dq.offerLast(i);

            res.add(nums[dq.peekFirst()]);

        }

        return res.stream().mapToInt(Integer::intValue).toArray();

    }

}