From c5686b417d24a1d999d34ba9cf29d85ce77fe602 Mon Sep 17 00:00:00 2001
From: Jasen <141482690+JasenWn@users.noreply.github.com>
Date: Tue, 27 Aug 2024 22:32:59 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B00503.=E4=B8=8B=E4=B8=80?=
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---
 problems/0503.下一个更大元素II.md | 30 +++++++++++++++++++++++-
 1 file changed, 29 insertions(+), 1 deletion(-)

diff --git a/problems/0503.下一个更大元素II.md b/problems/0503.下一个更大元素II.md
index 62066d85..5751bb91 100644
--- a/problems/0503.下一个更大元素II.md
+++ b/problems/0503.下一个更大元素II.md
@@ -168,6 +168,7 @@ class Solution {
 ```
 
 ### Python: 
+> 版本一：
 
 ```python
 class Solution:
@@ -181,6 +182,34 @@ class Solution:
             stack.append(i%len(nums))
         return dp
 ```
+
+> 版本二：针对版本一的优化
+
+```python3
+class Solution:
+    def nextGreaterElements(self, nums: List[int]) -> List[int]:
+        res = [-1] * len(nums)
+        stack = []
+        #第一次遍历nums
+        for i, num in enumerate(nums):   
+            while stack and num > nums[stack[-1]]:
+                res[stack[-1]] = num
+                stack.pop()
+            stack.append(i)
+        #此时stack仍有剩余，有部分数‘无下一个更大元素’待修正
+        #第二次遍历nums
+        for num in nums:
+            #一旦stack为空，就表明所有数都有下一个更大元素，可以返回结果    
+            if not stack:   
+                return res
+            while stack and num > nums[stack[-1]]:
+                res[stack[-1]] = num
+                stack.pop()
+            #不要将已经有下一个更大元素的数加入栈，这样会重复赋值，只需对第一次遍历剩余的数再尝试寻找下一个更大元素即可
+        #最后仍有部分最大数无法有下一个更大元素，返回结果
+        return res  
+```
+
 ### Go:
 
 ```go
@@ -203,7 +232,6 @@ func nextGreaterElements(nums []int) []int {
     return result
 }
 ```
-
 ### JavaScript:
 
 ```JS