leetcode-21合并两个有序链表

最新推荐文章于 2025-12-16 18:10:48 发布

原创最新推荐文章于 2025-12-16 18:10:48 发布 · 199 阅读

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文章标签：

#链表 #leetcode #算法 #java #nodejs

本文介绍了一种合并两个已排序链表的方法，通过迭代和递归两种方式实现，旨在帮助读者理解链表操作和排序合并的基本原理。

https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/he-bing-liang-ge-you-xu-lian-biao-by-leetcode-solu/

方法一：迭代

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        res = ListNode(None)
        node = res
        if not l1:
            return l2
        if not l2:
            return l1
        while l1 and l2:
            if l1.val < l2.val:
                node.next, l1 = l1, l1.next
            else:
                node.next, l2 = l2, l2.next
            node = node.next
        if l1:
            node.next = l1
        else:
            node.next = l2
        return res.next

class Solution:
    def mergeTwoLists(self, l1, l2):
        prehead = ListNode(-1)

        prev = prehead
        while l1 and l2:
            if l1.val <= l2.val:
                prev.next = l1
                l1 = l1.next
            else:
                prev.next = l2
                l2 = l2.next            
            prev = prev.next

        # 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
        prev.next = l1 if l1 is not None else l2

        return prehead.next

方法二：递归

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 is None:
            return l2
        elif l2 is None:
            return l1
        elif l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l2.next, l1)
            return l2