Python 实现 两个有序链表合并为一个新的有序链表

"""
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

示例：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

"""

链表结构：

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

class Solution:
    # 方法一 ：
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = ListNode(None)
        l3 = head
        while l1 and l2 :
            if l1.val < l2.val:
                # 重点： 时刻保持下一个 .next 结构是一个 ListNode
                l3.next = ListNode(l1.val)
                l1 = l1.next
            else:
                l3.next = ListNode(l2.val)
                l2 = l2.next
            l3 = l3.next
        if l1 == None:
            while l2:
                l3.next = ListNode(l2.val)
                l3 = l3.next
                l2 = l2.next
        else:
            while l1:
                l3.next = ListNode(l1.val)
                l3 = l3.next
                l1 = l1.next
        return head.next

 

   # 方法 二
   def mergeTwoLists2(self, l1: ListNode, l2: ListNode) -> ListNode:
        # 递归解法
        if not l1: return l2
        if not l2: return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

验证，设置输入、输出：

if __name__ == "__main__":
    s = Solution()
    #输入：    1->2->4
    #          1->3->4
    # 输出：1->1->2->3->4->4
    l1 = ListNode(1)
    tmp = ListNode(2)
    l1.next = tmp
    tmp.next = ListNode(4)
    l2 = ListNode(1)
    tmp = ListNode(3)
    l2.next = tmp
    tmp.next = ListNode(4)
    r = s.mergeTwoLists2(l1,l2)
    print(r.val)
    # 先判断是否为 None ， 再去打印
    while r.next != None:
        r = r.next
        print(r.val)