[LeetCode] 239. Sliding Window Maximum

轉自LeetCode

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]

Example 3:

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

Example 4:

Input: nums = [9,11], k = 2
Output: [11]

Example 5:

Input: nums = [4,-2], k = 2
Output: [4]

 

Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104
• 1 <= k <= nums.length

Solution

這題難在邊界變換的時候，還要記錄當下的最大值

關鍵點在用一個 queue 來記錄window size 裡最大值的 index

當 queue 的第一個值等於 index-k，代表第一個值已經在窗外了，可以移除

因為這個 queue 是要記錄最大值的 index

所以每次都要檢查，當下的值，是不是比前面已經在 queue 裡指向的值大

是的話，就要移除 queue 裡的值

最後，因為每次移動 window，都要記錄目前的最大值

所以當 index 大於等於 k-1 的時候，就代表 window 已經滿足條件，可以開始紀錄最大值了

kotlin(參考解答)

	class Solution {
	fun maxSlidingWindow(nums: IntArray, k: Int): IntArray {
	val ans = arrayListOf<Int>()
	val indexList = mutableListOf<Int>()
	for(i in nums.indices) {
	if(indexList.isNotEmpty() && indexList.first() == i - k) {
	indexList.removeAt(0)
	}
	while(indexList.isNotEmpty() && nums[indexList.last()] < nums[i]) {
	indexList.removeAt(indexList.lastIndex)
	}
	indexList.add(i)
	if(i >= k-1) {
	ans.add(nums[indexList.first()])
	}
	}
	return ans.toIntArray()
	}
	}

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