Update 0206.翻转链表.md

problems/0206.翻转链表.md

@@ -193,9 +193,9 @@ class Solution {
 }
 ```
 
-Python迭代法：
+Python
 ```python
-#双指针
+（版本一）双指针法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, val=0, next=None):
@@ -205,7 +205,7 @@ class Solution:
     def reverseList(self, head: ListNode) -> ListNode:
         cur = head   
         pre = None
-        while(cur!=None):
+        while cur:
             temp = cur.next # 保存一下 cur的下一个节点，因为接下来要改变cur->next
             cur.next = pre #反转
             #更新pre、cur指针
@@ -217,6 +217,7 @@ class Solution:
 Python递归法：
 
 ```python
+（版本二）递归法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, val=0, next=None):
@@ -224,36 +225,17 @@ Python递归法：
 #         self.next = next
 class Solution:
     def reverseList(self, head: ListNode) -> ListNode:
-        
+        return self.reverse(head, None)
-        def reverse(pre,cur):
+    def reverse(self, cur: ListNode, pre: ListNode) -> ListNode:
-            if not cur:
+        if cur == None:
             return pre
-                
+        temp = cur.next
-            tmp = cur.next
         cur.next = pre
-
+        return self.reverse(temp, cur)
-            return reverse(cur,tmp)
-        
-        return reverse(None,head)
 
 ```
 
-Python递归法从后向前：
 
-```python
-# Definition for singly-linked list.
-# class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
-class Solution:
-    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
-        if not head or not head.next: return head
-        p = self.reverseList(head.next)
-        head.next.next = head
-        head.next = None
-        return p
-```
 
 Go：