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.

Could you solve it in linear time?

## 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
}

``````

Sep 6, 2020