Algorithm Exploration_Maximum Sliding Window

Problem Description:

Given an array nums, a sliding window of size k moves from the leftmost side of the array to the rightmost side of the array. You can only see the k numbers in the sliding window. The sliding window only moves one position to the right at a time.

Returns the maximum value in the sliding window.

Advanced:

Can you solve this problem in linear time complexity?

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 

  The position of the sliding window Maximum value
      --------------- ----
[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

prompt:

1 <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4
1 <= k <= nums.length

Source: LeetCode
Link: https://leetcode-cn.com/problems/sliding-window-maximum

Solutions

Double pointer traversal

    /*
     *作者:赵星海
     *时间:2020/9/14 10:36
     *用途:滑动窗口最大值
     */
    public int[] maxSlidingWindow(int[] nums, int k) {
        //通过模拟k==1和k==3  得出结果的长度为nums.length+1-k
        int[] result = new int[nums.length + 1 - k];
        for (int i = 0; i < nums.length + 1 - k; i++) {
            int x = nums[i];
            for (int j = i + 1; j < (i + k); j++) {
                if (nums[j] > x) {
                    x = nums[j];
                }
            }
            result[i] = x;
        }
        return result;
    }