方法一、递归
List1或List2一开始就是空链表,那么不需要任何操作直接返回非空链表。否则我们就要判断List1或List2哪一个链表的头节点的值更小,然后递归地决定下一个添加到结果里的节点。如果两个链表有一个为空,递归结束。
Python3 代码题解
# 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 list1 is None:
return list2
elif list2 is None:
return list1
elif list1.val < list2.val:
list1.next = self.mergeTwoLists(list1.next, list2)
return list1
else:
list2.next = self.mergeTwoLists(list1, list2.next)
return list2
Source: LeetCode(The title reproduced in this blog is for personal study use only)
Be First to Comment