LeetCode31

m0_743849652024-08-22 10:47

206.反转链表

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

示例 1：

复制代码

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

复制代码

输入：head = [1,2]
输出：[2,1]

示例 3：

复制代码

输入：head = []
输出：[]

提示：

链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000

python 复制代码

# 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 is None:
            return None
        elif head.next is None:
            return head
        else:
            dummy = ListNode()
            p = head
            while p.next is not None:
                p = p.next
            dummy.next = p
            while 1:
                p = head
                while p.next.next is not None:
                    p = p.next
                p.next.next = p
                p.next = None
                if head.next is None:
                    return dummy.next

python 复制代码

# 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 is None:
            return None
        elif head.next is None:
            return head
        else:
            q = None
            p = head
            while p is not None:
                j = p.next
                p.next = q
                q = p
                p = j
            return q

总结

从后往前，不能遍历到最后一个，只能遍历到倒数第二个。

从前往后，需要三个指针，多一个存储下一个节点。