困難度:Easy
題目: 簡單來說呢!就是將兩個列表合併為一個排序列表。
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.
Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]
Input: list1 = [], list2 = []
Output: []
Input: list1 = [], list2 = [0]
Output: [0]
Inline code
has back-ticks around
it.
# 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, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
def findMin(n1,n2):
if n1 == None :
return n2
if n2 == None :
return n1
if n1.val < n2.val :
n1.next = findMin(n1.next , n2)
return n1
else:
n2.next = findMin(n1,n2.next)
return n2
return findMin(l1,l2)