From bec57550df7a01fccb7fe060ce569e499285a134 Mon Sep 17 00:00:00 2001
From: roylx <73628821+roylx@users.noreply.github.com>
Date: Thu, 10 Nov 2022 17:27:45 -0700
Subject: [PATCH] =?UTF-8?q?Update=200376.=E6=91=86=E5=8A=A8=E5=BA=8F?=
 =?UTF-8?q?=E5=88=97.md?=
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增加python动态规划优化版
---
 problems/0376.摆动序列.md | 19 +++++++++++++++++++
 1 file changed, 19 insertions(+)

diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md
index 214ff311..f9d3f97f 100644
--- a/problems/0376.摆动序列.md
+++ b/problems/0376.摆动序列.md
@@ -264,6 +264,25 @@ class Solution:
         return max(dp[-1][0], dp[-1][1])
 ```
 
+```python
+class Solution:
+    def wiggleMaxLength(self, nums: List[int]) -> int:
+        # up i作为波峰最长的序列长度
+	# down i作为波谷最长的序列长度
+        n = len(nums)
+	# 长度为0和1的直接返回长度
+	if n<2: return n
+        for i in range(1,n):
+            if nums[i]>nums[i-1]:
+	    # nums[i] 为波峰，1. 前面是波峰，up值不变，2. 前面是波谷，down值加1
+	    # 目前up值取两者的较大值(其实down+1即可，可以推理前一步down和up最多相差1，所以down+1>=up)
+                up = max(up, down+1)
+            elif nums[i]<nums[i-1]:
+	    # nums[i] 为波谷，1. 前面是波峰，up+1，2. 前面是波谷，down不变，取较大值
+                down = max(down, up+1)
+        return max(up, down)
+```
+
 ### Go 
 
 **贪心**