小刀初试 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==...
代码 1publicclassSolution {2publicListNode mergeTwoLists(ListNode l1, ListNode l2) {3//创建一个dummy头,从后面开始接4ListNode dummy =newListNode(0);5ListNode curr =dummy;6//依次比较拼接7while(l1 !=null&& l2 !=null){8if(l1.val <=l2.val){9curr.next =l1;10l1 =l1.next;11}else{12cu...
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. Seen this question in a real interview before? Yes 简单的归并排序,不说了,直接上代码:
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. 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 ...
21. 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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4...
21. Merge Two Sorted Lists题目描述: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.题目翻译将两个已排序的链表合并为一个新链表,新链表原链表的节点组成。
需要遍历 list2 中的全部 O(|list2|) 个结点 空间复杂度:O(1) 只需要维护常数个额外变量 代码(Python3) # 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, list...
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 head = pre while head1 and head2: if head1.val ...