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 >...
next = list2; } // 返回合并后的链表的头结点 head_pre.next } } 题目链接: Merge Two Sorted Lists : leetcode.com/problems/m 合并两个有序链表: leetcode-cn.com/problem LeetCode 日更第 52 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
*@returnListNode */functionmergeTwoLists($l1,$l2){if($l1==null) {return$l2; }if($l2==null) {return$l1; }if($l1->val <=$l2->val) {$l1->next =mergeTwoLists($l1->next,$l2);return$l1; }if($l2->val <=$l1->val) {$l2->next =mergeTwoLists($l2->next,$l1);return$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.解题思路:1)定义l,每次l指向两个链表中小的那个节点,定义head为头指针,定义ptr,将l依次连成链表2)两个链表均不为空,将其小的值赋给l3)将其中节点多的...
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode { if l1 == nil { return l2; }else if l2 == nil { return l1; }else if (l1.Val < l2.Val) { l1.Next = mergeTwoLists(l1.Next, l2); return l1; }else { l2.Next = mergeTwoLists(l1, l2.Next); ...
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 翻译: 合并两个有序链表并返回一个新的链表,新链表必须由前两个链表拼接...
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. 示例: 代码语言:javascript 复制 输入:1->2->4,1->3->4输出:1->1->2->3->4->4
Merge Two Sorted Lists 二、解题 两个有序链表的排序。 三、尝试与结果 classSolution(object):defmergeTwoLists(self,l1,l2):ret=ListNode(0)ifl1==None:returnl2ifl2==None:returnl1ifl1.val<l2.val:ret=l1 ret.next=self.mergeTwoLists(l1.next,l2)else:ret=l2 ...
varmergeTwoLists=function(l1,l2){letresult=null;//终止条件if(l1==null)returnl2;if(l2==null)returnl1;//判断数值大小递归if(l1.val<l2.val){result=l1;result.next=mergeTwoLists(l1.next,l2);}else{result=l2;result.next=mergeTwoLists(l2.next,l1);}//返回结果returnresult;}; ...
You are given the heads of two sorted linked listslist1andlist2. Merge the two lists into onesortedlist. The list should be made by splicing together the nodes of the first two lists. Return the head of the merged linked list.