如何合并两个有序链表

"""
已知两个链表head1和head2各自有序（例如升序排列），请把他们合并成一个链表，要求合并后的链表依然有序。
"""


class LNode:
    def __init__(self):
        self.data = None
        self.next = None


def ConstructList(start):
    """
    方法功能：构造链表
    :param start:
    :return:
    """

    i = start
    head = LNode()
    tmp = None
    cur = head
    while i < 7:
        tmp = LNode()
        tmp.data = i
        cur.next = tmp
        cur = tmp
        i += 2
    return head


def PrintList(head):
    cur = head.next
    while cur is not None:
        print(cur.data, end=' ')
        cur = cur.next


def Merge(head1, head2):
    """
    方法功能：合并两个升序排列的单链表
    :param head1: 链表1
    :param head2: 链表2
    :return: 合并后链表的头结点
    """

    if head1 is None or head1.next is None:
        return head2
    if head2 is None or head2.next is None:
        return head1

    cur1 = head1.next  # 用来遍历head1
    cur2 = head2.next  # 用来遍历head2
    head = None  # 合并后链表的头结点
    cur = None  # 合并后的链表在尾结点
    # 合并后链表的头结点为第一个结点元素最小的那个链表的头结点
    if cur1.data > cur2.data:
        head = head2
        cur = cur2
        cur2 = cur2.next
    else:
        head = head1
        cur = cur1
        cur1 = cur1.next
    # 每次找到链表剩余结点最小值对应的结点连接到合并后链表的尾部
    while cur1 is not None and cur2 is not None:
        if cur1.data < cur2.data:
            cur.next = cur1
            cur = cur1
            cur1 = cur1.next
        else:
            cur.next = cur2
            cur = cur2
            cur2 = cur2.next
    # 当遍历完一个链表后把另外一个链表剩余结点连接到合并后的链表后边
    if cur1 is not None:
        cur.next = cur1
    if cur2 is not None:
        cur.next = cur2
    return head


if __name__ == '__main__':
    head1 = ConstructList(1)
    head2 = ConstructList(2)
    print('head1:')
    PrintList(head1)

    print('\nhead2:')
    PrintList(head2)

    print('\n合并后的链表')
    head = Merge(head1, head2)
    PrintList(head)

运行j结果如下：
head1:
1 3 5
head2:
2 4 6
合并后的链表
1 2 3 4 5 6