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 ...
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->4Output:1->1->2->3->4->4 因为没有空间要求,所以想到ListNode*head = new ListNode(INT_MIN);重新定义一...
Quest: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. 合并两个有序数列 题目给出的原型类 publicclassListNode {intval; ListNode next; ListNode(intx) { val =x; } }publicListNode mergeTwoLists(L...
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...
* Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */struct ListNode*mergeTwoLists(struct ListNode*l1,struct ListNode*l2){} 通过函数定义可以看出,mergeTwoLists 函数的参数是两个单向链表,然后返回值也是一个链表,我们要做的就是把两个有序的链表合...
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. 题目大意:合并两个有序的链表 思路:通过比较两个链表的节点大小,采用尾插法建立链表。
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. 思路: easy 。 算法: AI检测代码解析 1. public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ...
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:
链接:https://leetcode.cn/problems/merge-two-sorted-lists/solution/he-bing-liang-ge-you-xu-lian-biao-by-leetcode-solu/来源:力扣(LeetCode) 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。 可以看到,有一个链表是null的情况下,直接return了非空链表 ...
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. 翻译:合并两个排好序的链列,并将其作为新链表返回。新链表应通过将前两个列表的节点拼接在一起。