leetcode专题训练 206. Reverse Linked List

反转链表的问题，prev存储当前p的前一个结点，也就是反转之后当前p应该指向的结点，p1存储当前p的下一个结点，用于p的更新。反转时，核心语句为p.next=prev，之后依次更新prev和p。当p为None时反转完毕，此时的prev就是新链表的头。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        prev = None
        p = head
        while p is not None:
            p1 = p.next
            p.next = prev
            prev = p
            p = p1
            
        return prev