206. 反转链表
题解:
定义两个指针prev和cur,分别指向当前节点的前一个节点和当前节点,初始时,prev为None,cur为head,在遍历的过程中,我们需要创建一个临时指针用来指向cur.next,因为cur指向prev之后,就会和cur.next断链了。遍历过程是,每次将cur的next指针执行prev,然后将prev和cur分别向后移动一位,当遍历到尾部时,链表就被反转了,反转后的链表头节点是prev.
# 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]:prev = Nonecur = headwhile cur:tmp = cur.nextcur.next = prevprev = curcur = tmpreturn prev