[leetcode] 21. Merge Two Sorted Lists @ python_21. merge two sorted lists(python)-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/86183393

本文详细介绍了两种合并两个有序链表的方法：迭代法和递归法。迭代法通过建立虚拟节点，遍历两个链表并将较小节点连接到当前节点；递归法则利用分治策略，选择较小值作为新链表的头部，再递归处理剩余部分。

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原题

Merge Two Sorted Lists
Easy

1758

240

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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

解法1

迭代法, 建立虚拟变量dummy, 遍历L1和L2, 每次将较小的节点作为当前指针p的下一个节点, 当某个链表遍历完了, p指向另一个链表, 最后返回头部节点.
Time: O(n)
Space: O(1)
代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

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

解法2

递归, 先找到base case: 当某个链表为空时, 返回另一链表. 然后使用分治法, 如果L1的值更小, 那么头部节点是L1, L1的下一个节点是递归后的节点, 反之亦然.
Time: O(n)
Space: O(1)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if l1 is None or l2 is None:
            return l2 or 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