0239. Sliding Window Maximum

239. Sliding Window Maximum #

题目 #

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?

题目大意 #

给定一个数组 nums,有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口 k 内的数字。滑动窗口每次只向右移动一位。返回滑动窗口最大值。

解题思路 #

  • 给定一个数组和一个窗口为 K 的窗口,当窗口从数组的左边滑动到数组右边的时候,输出每次移动窗口以后,在窗口内的最大值。
  • 这道题最暴力的方法就是 2 层循环,时间复杂度 O(n * K)。
  • 另一种思路是用优先队列,每次窗口以后的时候都向优先队列里面新增一个节点,并删除一个节点。时间复杂度是 O(n * log n)
  • 最优的解法是用双端队列,队列的一边永远都存的是窗口的最大值,队列的另外一个边存的是比最大值小的值。队列中最大值左边的所有值都出队。在保证了双端队列的一边即是最大值以后,时间复杂度是 O(n),空间复杂度是 O(K)

代码 #


package leetcode

// 解法一 暴力解法 O(nk)
func maxSlidingWindow1(a []int, k int) []int {
	res := make([]int, 0, k)
	n := len(a)
	if n == 0 {
		return []int{}
	}
	for i := 0; i <= n-k; i++ {
		max := a[i]
		for j := 1; j < k; j++ {
			if max < a[i+j] {
				max = a[i+j]
			}
		}
		res = append(res, max)
	}

	return res
}

// 解法二 双端队列 Deque
func maxSlidingWindow(nums []int, k int) []int {
	if len(nums) == 0 || len(nums) < k {
		return make([]int, 0)
	}
	window := make([]int, 0, k) // store the index of nums
	result := make([]int, 0, len(nums)-k+1)
	for i, v := range nums { // if the left-most index is out of window, remove it
		if i >= k && window[0] <= i-k {
			window = window[1:len(window)]
		}
		for len(window) > 0 && nums[window[len(window)-1]] < v { // maintain window
			window = window[0 : len(window)-1]
		}
		window = append(window, i) // store the index of nums
		if i >= k-1 {
			result = append(result, nums[window[0]]) // the left-most is the index of max value in nums
		}
	}
	return result
}


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