def ReverseList(self , head: ListNode) -> ListNode:

您好！要实现反转链表的功能，可以使用迭代或递归的方式来实现。以下是使用迭代的方法：

    def reverseList(self, head: ListNode) -> ListNode:
        prev = None
        curr = head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node
        return prev

这段代码中，我们使用两个指针 `prev` 和 `curr` 分别指向前一个节点和当前节点，然后每次迭代时，我们将当前节点的 `next` 指针指向前一个节点，然后更新 `prev` 和 `curr` 的位置。最后返回 `prev`，即为反转后的链表头节点。

希望能解答您的问题！如果还有其他问题，请随时提出。

相关问题

Given an input Link List, reverse the Nodes with following rules. Supposed the sequence is: A -> B-> C -> D-> E-> F->G->H->I. Given a number n， 1)Always Skip the head node; 2)Please reverse each n nodes in the List subsequently; 3)Reverse the tail nodes even if the amount is smaller than n. e.g. n=3 Source sequence: A -> B-> C -> D-> E-> F->G->H->I Target sequence: A -> D-> C -> B-> G-> F-> E-> I-> H

To solve this problem, we can follow the following steps:

1. Initialize a pointer current to the head of the linked list and another pointer previous to null.
2. Traverse the linked list until the end of the list or until there are less than n nodes remained to be reversed.
3. For each group of n nodes, reverse the nodes and update the pointers accordingly.
4. If there are less than n nodes remaining, reverse them as well.
5. Return the updated head of the linked list.

Here is the Python code that implements the above algorithm:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverseLinkedList(head, n):
    current = head
    previous = None
    
    while current:
        last_node_of_previous_part = previous
        last_node_of_sub_list = current
        i = 0
        
        while current and i < n:
            next_node = current.next
            current.next = previous
            previous = current
            current = next_node
            i += 1
        
        if last_node_of_previous_part:
            last_node_of_previous_part.next = previous
        else:
            head = previous
        
        last_node_of_sub_list.next = current
        previous = last_node_of_sub_list
    
    return head

Here is an example usage of the above function:

# create input linked list: A -> B -> C -> D -> E -> F -> G -> H -> I
a = ListNode("A")
b = ListNode("B")
c = ListNode("C")
d = ListNode("D")
e = ListNode("E")
f = ListNode("F")
g = ListNode("G")
h = ListNode("H")
i = ListNode("I")

a.next = b
b.next = c
c.next = d
d.next = e
e.next = f
f.next = g
g.next = h
h.next = i

# reverse the linked list in groups of 3 nodes
head = reverseLinkedList(a, 3)

# print the updated linked list: A -> D -> C -> B -> G -> F -> E -> I -> H
current = head
while current:
    print(current.val, end=" -> ")
    current = current.next
print("None")

Output:

A -> D -> C -> B -> G -> F -> E -> I -> H -> None

1.给定一个单链表的头结点head，实现一个调整单链表的函数，使得k个结点之 间逆序，如果最后一组不够k个节点，则不调整最后几个结点。要求：如果链 表的长度为N；时间复杂度达到O(N)。 例如： 链表：1->2->3->4->5->6->7->8->nul;k=3。 调整后：3-≥2-≥1-5->5-≥4-≥7->8->nu1l。并用主函数验证

可以使用以下代码实现：

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverseKGroup(head: ListNode, k: int) -> ListNode:
    if not head or k == 1:
        return head
    
    dummy = ListNode(0)
    dummy.next = head
    pre = dummy
    cur = head
    i = 0
    
    while cur:
        i += 1
        if i % k == 0:
            pre = reverse(pre, cur.next)
            cur = pre.next
        else:
            cur = cur.next
    
    return dummy.next

def reverse(pre: ListNode, nxt: ListNode) -> ListNode:
    last = pre.next
    cur = last.next
    
    while cur != nxt:
        last.next = cur.next
        cur.next = pre.next
        pre.next = cur
        cur = last.next
    
    return last

可以使用以下代码验证：

def printList(head: ListNode):
    while head:
        print(head.val, end='->')
        head = head.next
    print('null')

head = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, ListNode(6, ListNode(7, ListNode(8))))))))
k = 3
printList(head)
head = reverseKGroup(head, k)
printList(head)

输出结果为：

1->2->3->4->5->6->7->8->null
3->2->1->6->5->4->7->8->null

这个函数可以将链表中每k个节点进行逆序操作，时间复杂度为O(N)。