LeetCode 솔루션

[Easy - wk2 - Q1] 21. Merge Two Sorted Lists

Chloe 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.05.08 20:08

7,385 조회
2 댓글
0 추천
0 비추천
목록 게시글 쓰기

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.

#leetcode, #문제풀이, #easy, #amazon, #microsoft, #facebook, #adobe, #oracle, #uber, #google, #bloomerg, #shopee, #accenture, #linked list, #recursion

링크

https://leetcode.com/problems/merge-two-sorted-lists/ 2690 회 연결

익명
작성일 2022.05.09 15:15

class Solution:
    def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(0)
        l3 = dummy
        while l1 and l2:
            if l1.val < l2.val:
                l3.next = l1
                l1 = l1.next
            else:
                l3.next = l2
                l2 = l2.next
            l3 = l3.next
            
        if l1:
            l3.next = l1
        if l2:
            l3.next = l2
            
        return dummy.next

익명
작성일 2022.05.11 19:17

Python

Runtime: 57 ms, faster than 36.18% of Python3 online submissions for Merge Two Sorted Lists.
Memory Usage: 13.8 MB, less than 80.22% of Python3 online submissions for Merge Two Sorted Lists.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        if ( not list1) or (list2 and list1.val > list2.val):
            list1, list2 = list2, list1
        if list1:
            list1.next = self.mergeTwoLists(list1.next, list2)
        return list1