[LeetCode] 21. Merge Two Sorted Lists_Easy tag: Linked List

最新推荐文章于 2025-09-06 22:33:46 发布

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最新推荐文章于 2025-09-06 22:33:46 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/Johnsonxiong/p/10801797.html

博客围绕合并两个有序链表展开，提出使用dummy node，依次比较l1和l2节点大小，当一方为None时将另一方加在总链表末尾的思路，并给出示例，代码转载自相关博客。

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

这个题目思路就是用dummy node，然后依次判断是l1 小还是l2小，最后当一方是None的时候将另一方加在总list的最后即可。

Code

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        dummy = ListNode(0)
        head = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                head.next = l1
                l1 = l1.next
            else:
                head.next = l2
                l2 = l2.next
            head = head.next
        head.next = l1 if l1 else l2
        return dummy.next