leetCode 206 反转链表（头插法+ 两种递归）

递归

1. # 优先处理尾部的链表，此时需要返回子链表的头节点和尾节点
   # 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 head:
               head, _ = self.func(head)
           return head
   
       def func(self, head):
           if not head.next:
               return head, head
           cur = head
           sonh, sont = self.func(cur.next)
           sont.next, cur.next = cur, None
           return sonh, cur
   
   

2. # 优先处理头部链表，此时参数中需要指明前节点pre和当前节点p
   class Solution:
       def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
           return self.func(None, head)
   
       def func(self, pre, p):
           if not p:
               return pre
           nxt, p.next = p.next, pre
           return self.func(p, nxt)
   

迭代

1. # 采用头插法, 多元赋值时会先把右边的计算出来，然后再进行赋值
   
   # 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]:
           dmyn = ListNode()
           dmyn.next, p = head, head
           while p and p.next:
               p.next.next, dmyn.next, p.next = dmyn.next, p.next, p.next.next
   
           
           return dmyn.next