# 反转一个单链表。
#
# 示例:
#
# 输入: 1->2->3->4->5->NULL
# 输出: 5->4->3->2->1->NULL
# 进阶:
# 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

# 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, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre


if __name__ == '__main__':
    l1 = ListNode(1)
    l2 = ListNode(2)
    l3 = ListNode(3)
    l4 = ListNode(4)
    l1.next = l2
    l2.next = l3
    l3.next = l4
    l4.next = None
    s = Solution()
    pre = s.reverseList(l1)
    while pre is not None:
        print(pre.val)
        pre = pre.next
1532652ec5042e1fa0224239a27d3f6dd1209a02
<mask token>


class Solution:

    def reverseList(self, head: ListNode) ->ListNode:
        cur, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre


<mask token>

class ListNode:
    <mask token>


class Solution:

    def reverseList(self, head: ListNode) ->ListNode:
        cur, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre


<mask token>

class ListNode:

    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:

    def reverseList(self, head: ListNode) ->ListNode:
        cur, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre


<mask token>

class ListNode:

    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:

    def reverseList(self, head: ListNode) ->ListNode:
        cur, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre


if __name__ == '__main__':
    l1 = ListNode(1)
    l2 = ListNode(2)
    l3 = ListNode(3)
    l4 = ListNode(4)
    l1.next = l2
    l2.next = l3
    l3.next = l4
    l4.next = None
    s = Solution()
    pre = s.reverseList(l1)
    while pre is not None:
        print(pre.val)
        pre = pre.next

# 反转一个单链表。
#
# 示例:
#
# 输入: 1->2->3->4->5->NULL
# 输出: 5->4->3->2->1->NULL
# 进阶:
# 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

# 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, pre = head, None
        while cur:
            cur.next, pre, cur = pre, cur, cur.next
        return pre


if __name__ == '__main__':
    l1 = ListNode(1)
    l2 = ListNode(2)
    l3 = ListNode(3)
    l4 = ListNode(4)
    l1.next = l2
    l2.next = l3
    l3.next = l4
    l4.next = None
    s = Solution()
    pre = s.reverseList(l1)
    while pre is not None:
        print(pre.val)
        pre = pre.next