LeetCode 反转链表

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

按照过程依次翻转。

# 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:
        cur_prt = head
        done_prt = None
        if cur_prt == None:
            return head
        wait_prt = cur_prt.next 
        
        while(wait_prt):
            cur_prt.next = done_prt
            done_prt = cur_prt
            cur_prt = wait_prt
            wait_prt = wait_prt.next
            
        cur_prt.next = done_prt
        head = cur_prt
        return head

还有一种方案就是根据现有的函数进行递归迭代，因为每部都是在上一步翻转的基础上进一步的操作。因为已知reverseList(head)返回翻转后的链表，所以返回reverseList(heed.next)的结果，并对 head.next 进行更新

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:

        if head == None or head.next == None:
            return head
        ### 递归方法
        node = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return node