题解 | #反转链表#

class Solution:
    def ReverseList(self , head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        cur = head
        pre = None
        while cur!=None:
            tmp = cur.next
            cur.next = pre 
            pre = cur
            cur = tmp
        return pre
反转列表需要一个pre一个tmp
先要记住下一个位置