设计把L1中与L2中数据相同的连续结点顺序完全

可以使用双指针的方法，首先将两个链表分别按照升序排列，然后分别从两个链表的头结点开始，比较当前结点的值大小，将值较小的结点插入到结果链表中，并将指针后移一位。如果两个结点的值相等，则将它们依次插入到结果链表中，直到有一个链表遍历完为止。最后，将未遍历完的链表中的剩余结点全部插入到结果链表的末尾即可。

具体实现可以参考下面的代码：

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

def mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode:
    # 将两个链表按升序排列
    dummy = ListNode(-1)
    cur = dummy
    while l1 and l2:
        if l1.val <= l2.val:
            cur.next = l1
            l1 = l1.next
        else:
            cur.next = l2
            l2 = l2.next
        cur = cur.next
    cur.next = l1 if l1 else l2
    return dummy.next

def mergeDuplicateNodes(head: ListNode) -> ListNode:
    if not head or not head.next:
        return head
    # 将链表拆分成两个部分
    slow, fast = head, head.next
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    l1, l2 = head, slow.next
    slow.next = None
    # 对两个链表进行排序合并
    l1_sorted = mergeSort(l1)
    l2_sorted = mergeSort(l2)
    # 将两个链表中相同的连续结点顺序完全
    dummy = ListNode(-1)
    cur = dummy
    while l1_sorted and l2_sorted:
        if l1_sorted.val < l2_sorted.val:
            cur.next = l1_sorted
            l1_sorted = l1_sorted.next
        elif l1_sorted.val > l2_sorted.val:
            cur.next = l2_sorted
            l2_sorted = l2_sorted.next
        else:
            cur1, cur2 = l1_sorted, l2_sorted
            while cur1 and cur2 and cur1.val == cur2.val:
                cur1 = cur1.next
                cur2 = cur2.next
            if not cur1 or (cur2 and cur1.val > cur2.val):
                cur.next = l2_sorted
                l2_sorted = cur2
            else:
                cur.next = l1_sorted
                l1_sorted = cur1
        cur = cur.next
    cur.next = l1_sorted if l1_sorted else l2_sorted
    return dummy.next

def mergeSort(head: ListNode) -> ListNode:
    if not head or not head.next:
        return head
    # 将链表拆分成两个部分
    slow, fast = head, head.next
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    l1, l2 = head, slow.next
    slow.next = None
    # 对两个链表进行归并排序
    l1_sorted = mergeSort(l1)
    l2_sorted = mergeSort(l2)
    # 合并两个排序后的链表
    return mergeTwoLists(l1_sorted, l2_sorted)

该算法的时间复杂度为 $O(n \log n)$，其中 $n$ 是链表的长度。