要保持原始顺序的情况下翻转链表，可以使用迭代或递归的方法来解决。以下是两种方法的代码示例：

1. 迭代方法:

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

def reverseLinkedList(head):
    prev = None
    curr = head
    
    while curr:
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node
    
    return prev

2. 递归方法:

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

def reverseLinkedList(head):
    if not head or not head.next:
        return head
    
    reversed_head = reverseLinkedList(head.next)
    head.next.next = head
    head.next = None
    
    return reversed_head

上述两种方法的时间复杂度都是O(n)，其中n是链表的长度。迭代方法使用了三个指针来逐个翻转节点，而递归方法则通过递归调用来实现翻转。