本身有序

@ycyzharry: 我写了一段代码，满足你的需求，不过没有测其它的极端情况，仅供参考：

class Node(object):
    def __init__(self, val):
        self.__val = val
        self.__next = None

    @property
    def val(self):
        return self.__val

    @property
    def next(self):
        return self.__next

    @next.setter
    def next(self, next):
        self.__next = next

    def __str__(self):
        display = ''
        node = self
        while node:
            if display == '':
                display = str(node.val)
            else:
                display = display + '->' + str(node.val)
            node = node.next
        return display

def merge(node1, node2):
    ''' 合并两个链表，保持有序性'''
    
    newnode = None
    
    if node1 == None:
        newnode = node2
    elif node2 == None:
        newnode = node1
    else:
        if node1.val < node2.val:
            newnode = Node(node1.val)           
            node1 = node1.next
        else:
            newnode = Node(node2.val)
            node2 = node2.next

        currentnode = newnode

        while True:
            if node1 == None and node2 == None:
                break
            else:
                if node1 == None:
                    currentnode.next = Node(node2.val)
                    currentnode = currentnode.next
                    node2 = node2.next
                elif node2 == None:
                    currentnode.next = Node(node1.val)
                    currentnode = currentnode.next
                    node1 = node1.next
                else:
                    if node1.val < node2.val:
                        currentnode.next = Node(node1.val)
                        currentnode = currentnode.next
                        node1 = node1.next
                    else:
                        currentnode.next = Node(node2.val)
                        currentnode = currentnode.next
                        node2 = node2.next
    return newnode

if __name__ == "__main__":
    
    node1 = Node(1)
    node2 = Node(2)
    node3 = Node(3)
    node4 = Node(4)
    node5 = Node(5)
    node8 = Node(8)

    node1.next = node3
    node3.next = node4

    node2.next = node5
    node5.next = node8

    print(node1) # 1->3->4
    print(node2) # 2->5->8

    newnode = merge(node1, node2)
    print(newnode) # 1->2->3->4->5->8

我没猜错的话：你肯定是会写的，只是为了赶时间才提了这个问题，想让别人给你写。我偏偏不用C#写，让你再重写一遍，哈哈哈，我也是闲的蛋疼啊。

@会长: 谢谢

java版本：

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode listNode = new ListNode(0);
    ListNode firstNode = listNode;

    while (l1 != null && l2 != null) {
        if (l1.val <= l2.val) {
            listNode.next = l1;
            l1 = l1.next;
        } else {
            listNode.next = l2;
            l2 = l2.next;
        }
        listNode = listNode.next;
    }
    while (l1 != null) {
        listNode.next = l1;
        l1 = l1.next;
        listNode = listNode.next;
    }
    while (l2 != null) {
        listNode.next = l2;
        l2 = l2.next;
        listNode = listNode.next;
    }
    return firstNode.next;
}

思路：
第一个while循环，将 l1 和 l2 进行比较，谁小谁就合并到 listNode，直到 l1 或者 l2 为空
第二个while循环和第三个while循环，将 l1 或者 l2 剩下的节点合并到 listNode
最后返回的是 firstNode.next //头节点是不需要的
其他方案可参考：
java实现两个有序单链表合并