feat($239): 增加python3解法

Author: azl397985856

problems/239.sliding-window-maximum.md

@@ -33,12 +33,9 @@ Could you solve it in linear time?
 符合直觉的想法是直接遍历 nums, 然后然后用一个变量 slideWindow 去承载 k 个元素，
 然后对 slideWindow 求最大值，这是可以的，时间复杂度是 O(n \* k).代码如下：
 
+JavaScript:
+
 ```js
-/**
- * @param {number[]} nums
- * @param {number} k
- * @return {number[]}
- */
 var maxSlidingWindow = function(nums, k) {
   // bad 时间复杂度O(n * k)
   if (nums.length === 0 || k === 0) return [];
@@ -54,6 +51,17 @@ var maxSlidingWindow = function(nums, k) {
   return ret;
 };
 ```
+Python3:
+
+```python
+class Solution:
+    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
+        if k == 0: return []
+        res = []
+        for r in range(k - 1, len(nums)):
+            res.append(max(nums[r - k + 1:r + 1]))
+        return res
+```
 
 但是如果真的是这样，这道题也不会是 hard 吧？这道题有一个 follow up，要求你用线性的时间去完成。
 我们可以用双端队列来完成，思路是用一个双端队列来保存`接下来的滑动窗口可能成为最大值的数`。具体做法：
@@ -81,54 +89,10 @@ var maxSlidingWindow = function(nums, k) {
 
 ## 代码
 
+
+JavaScript:
+
 ```js
-/*
- * @lc app=leetcode id=239 lang=javascript
- *
- * [239] Sliding Window Maximum
- *
- * https://leetcode.com/problems/sliding-window-maximum/description/
- *
- * algorithms
- * Hard (37.22%)
- * Total Accepted:    150.8K
- * Total Submissions: 399.5K
- * Testcase Example:  '[1,3,-1,-3,5,3,6,7]\n3'
- *
- * 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?
- */
-/**
- * @param {number[]} nums
- * @param {number} k
- * @return {number[]}
- */
 var maxSlidingWindow = function(nums, k) {
   // 双端队列优化时间复杂度, 时间复杂度O(n)
   const deque = []; // 存放在接下来的滑动窗口可能成为最大值的数
@@ -153,6 +117,24 @@ var maxSlidingWindow = function(nums, k) {
 };
 ```
 
+Python3:
+
+```python
+class Solution:
+    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
+        deque, res, n = [], [], len(nums)
+        for i in range(n):
+            while deque and deque[0] < i - k + 1:
+                deque.pop(0)
+            while deque and nums[i] > nums[deque[-1]]:
+                deque.pop(-1)
+            deque.append(i)
+            if i >= k - 1: res.append(nums[deque[0]])
+        return res
+
+
+```
+
 ## 扩展
 
 ### 为什么用双端队列