1 題目描述

2 解題思路

2.1 哨兵節點+循環

# 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: ListNode, l2: ListNode) -> ListNode:tmp=ListNode(0)#哨兵節點ret=tmpwhile(l1!=None and l2!=None): #如果l1和l2都有值得情況下if(l1.val<=l2.val):tmp.next=ListNode(l1.val)tmp=tmp.nextl1=l1.nextelse:tmp.next=ListNode(l2.val)tmp=tmp.nextl2=l2.nextif(l1!=None):tmp.next=l1if(l2!=None):tmp.next=l2 #如果l1和l2還有一個有值return(ret.next)

2.2 迭代

# 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: ListNode, l2: ListNode) -> ListNode:if(l1==None and l2==None):return Noneelif(l2==None):return l1elif(l1==None):return l2elif(l1.val>=l2.val):tmp=ListNode(l2.val)tmp.next=self.mergeTwoLists(l1,l2.next)return tmpelse:tmp=ListNode(l1.val)tmp.next=self.mergeTwoLists(l1.next,l2)return tmp

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總結

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