21.merge-two-sorted-lists

Statement

简体中文English

Metadata

• Link: 合并两个有序链表
• Difficulty: Easy
• Tag: 递归 链表

将两个升序链表合并为一个新的 升序 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 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

 

提示：

• 两个链表的节点数目范围是 [0, 50]
• -100 <= Node.val <= 100
• l1 和 l2 均按 非递减顺序 排列

Metadata

• Link: Merge Two Sorted Lists
• Difficulty: Easy
• Tag: Recursion Linked List

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

 

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

 

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

Solution

Python 3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
from typing import Optional


class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        rt = ListNode()
        res = rt
        while True:
            if not list1 and not list2:
                break
            if not list1:
                rt.next = list2
                list2 = list2.next
            elif not list2:
                rt.next = list1
                list1 = list1.next
            else:
                if list1.val < list2.val:
                    rt.next = list1
                    list1 = list1.next
                else:
                    rt.next = list2
                    list2 = list2.next

            rt = rt.next
        return res.next

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最后更新: October 11, 2023