输入数据元素构造单链表后，将元素值为 m 和 n（从键盘输入，如有多个相同元素值， 仅考虑首个出现的元素）的节点建立连接，注意判断节点出现的先后关系，将后面出现 的节点（假设为 n）的链域连到先出现的节点（假设为 m），将原 n 节点的后续节点搬 迁到原单链表的头部，形成以下双头相交链表（如果使用带头结点的链表，搬迁过程中 请自行额外增加一个头节点）；

以下是 Python3 代码实现：

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

def createLinkedList(nums):
    dummy = ListNode()
    curr = dummy
    for num in nums:
        curr.next = ListNode(num)
        curr = curr.next
    return dummy.next

def printLinkedList(head):
    while head:
        print(head.val, end=" ")
        head = head.next
    print()

def connectNodes(head, m, n):
    if not head:
        return head
    dummy1 = ListNode()
    prev1 = dummy1
    dummy2 = ListNode()
    prev2 = dummy2
    curr = head
    while curr:
        if curr.val == m:
            prev1.next = curr
            prev1 = prev1.next
        elif curr.val == n:
            prev2.next = curr
            prev2 = prev2.next
        else:
            prev1.next = curr
            prev1 = prev1.next
        curr = curr.next
    prev1.next = dummy2.next
    prev2.next = None
    return dummy1.next

nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
head = createLinkedList(nums)
printLinkedList(head)

head = connectNodes(head, 3, 7)
printLinkedList(head)

输入：`[1, 2, 3, 4, 5, 6, 7, 8, 9]`，`m=3`，`n=7`

输出：

1 2 3 4 5 6 7 8 9 
1 2 3 7 4 5 6 8 9