next return list class Solution: def mergeTwoLists(self, head1, head2): if not head1: # 如果head1为空,直接返回head2 return head2 if not head2: # 如果head2为空,直接返回head1 return head1 pre = ListNode(0) # 使用哑结点简化操作
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. » Solve this problem [Thoughts] 简单的实现,也没什么可说的。 [Code] 1:ListNode*mergeTwoLists(ListNode*l1,ListNode*l2){ 2:if(l1==NULL)retur...
题目:Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lis 相对还是简单的,直接合并。 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) ...
# 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]: # 使用一个哨兵结点 head_pre ...
Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. C++代码如下: #include<iostream>#include<new>usingnamespacestd;//Definition for singly-linked list.structListNode...
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 1. 2. # Definition for singly-linked list. ...
The output list should also be sorted from lowest to highest. Your algorithm should run in linear time on the length of the output list. 2 数据结构有序链表问题。 Write a function to merge two sorted linked lists. The input lists have their elements in sorted order, from lowest to highest...
Write a function to merge two sorted linked lists. T he input lists have their elements in sorted order, from lowest to highest. T he outputlist should also be sorted from lowest to highest. Your algorithm should run in linear time on the length of the output list....
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Solution1:Iterative遍历 思路:持续遍历比较,将小的relink到新的结果list。最后将多余的link一下 ...
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 题意 将两个有序链表合并 ...