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 = list1; } else { // 如果 list1 没有结点,表明 list2 已遍历完成, // 则将 list2 直接放在 tail 后面 tail.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/...
publicclassEasy_21_MergeTwoSortedList{publicstaticvoidmain(String[] args){Easy_21_MergeTwoSortedListinstance=newEasy_21_MergeTwoSortedList();ListNodel1=newListNode(1);ListNodel2=newListNode(2);ListNodel3=newListNode(4); l1.next = l2; l2.next = l3; System.out.println(instance.listNodeToString(...
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...
Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. Example 1: Input: l1 = [1,2,4], l2 = [1,3,4] Output: [1,1,2,3,4,4] Example 2: ...
LeetCode第21题 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 ...
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 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 Solve: ▉ 算法思路 ...
Return the head of the merged linked list. implSolution{pubfnmerge_two_lists(l1:Option<Box<ListNode>>,l2:Option<Box<ListNode>>,)->Option<Box<ListNode>>{match(l1,l2){(None,None)=>None,(Some(l1),None)=>Some(l1),(None,Some(l2))=>Some(l2),(Some(mutl1),Some(mutl2))=>{ifl1.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. 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。