【LeetCode】Merge Two Sorted Lists

最新推荐文章于 2025-04-27 14:03:08 发布

VegB

最新推荐文章于 2025-04-27 14:03:08 发布

阅读量266

点赞数

CC 4.0 BY-SA版权

分类专栏： Python LeetCode 文章标签： leetcode 合并

本文链接：https://blog.youkuaiyun.com/daphne566/article/details/54845951

Python 同时被 2 个专栏收录

14 篇文章

订阅专栏

LeetCode

6 篇文章

订阅专栏

本文介绍了一种高效的方法来合并两个已排序的链表。通过直接利用原链表的节点，避免了创建新链表所需的额外空间。提供两种实现方式：原始版与优化版，并解释了如何将剩余链表部分简便地链接到结果链表末尾。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Merge Two Sorted Lists

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.

Code

原始版 72ms

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        head = ListNode(0)
        tmp = head
        while l1 != None and l2 != None:
            if l1.val < l2.val:
                tmp.next = ListNode(l1.val)
                l1 = l1.next
            else:
                tmp.next = ListNode(l2.val)
                l2 = l2.next
            tmp = tmp.next
        if l1 != None:
            l = l1
        else:
            l = l2
        while l != None:
            tmp.next = ListNode(l.val)
            tmp = tmp.next
            l = l.next
        return head.next

2. 优化版 52ms

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        head = ListNode(0)
        tmp = head
        while l1 != None and l2 != None:
            if l1.val < l2.val:
                tmp.next = l1
                l1 = l1.next
            else:
                tmp.next = l2
                l2 = l2.next
            tmp = tmp.next
        if l1 != None:
            tmp.next = l1
        else:
            tmp.next = l2
        return head.next