改进mergeTwoLists方法,以在开始时检查空链表。 class ListNode: def __init__(self, x): self.val = x self.next = None # 改进后的将给出的数组转换为链表的函数 def linkedlist(list): if not list: # 检查列表是否为空 return None # 空列表返回None head = ListNode(list[0]) cur = head for...
You are given the heads of two sorted linked lists list1 and list2. Merge the two lists in a onesortedlist. The list should be made by splicing together the nodes of the first two lists. Returnthe head of the merged linked list. Example 1: Input: list1 = [1,2,4], list2 = [1...
然后我们再顺便复习下,怎么merge two linked list,代码如下: 1publicListNode mergeTwoLists(ListNode leftlist, ListNode rightlist){ 2if(rightlist ==null) 3returnleftlist; 4if(leftlist ==null) 5returnrightlist; 6 7ListNode fakehead =newListNode(-1); 8ListNode ptr = fakehead; 9while(rightlist!
*@returnListNode */functionmergeTwoLists($list1,$list2){$res=null;$this->merge($list1,$list2,$res);return$res; }functionmerge($l1,$l2, &$node){if(null==$l1&&null==$l2)return;$node=newListNode();if(null==$l1) {$node->val =$l2->val;$this->merge(null,$l2->next,$node->...
* Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode { if l1 == nil { return l2; }else if l2 == nil { return l1; }else if (l1.Val < l2.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->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...
package chapter_2_listproblem;publicclassProblem_19_MergeTwoLinkedLists{publicstaticclassNode{publicintvalue;publicNodenext;publicNode(intdata){this.value=data;}}publicstaticNodemerge(Nodehead1,Nodehead2){if(head1==null||head2==null){returnhead1!=null?head1:head2;}Nodehead=head1.value<head2....
l1.next=self.mergeTwoLists(l1.next,l2)returnl1orl2 以上两个方法使用了迭代的思想,LeetCode处理linked list更常用的方法是指针。在新list前加入一个dummy head。然后使用一个指针来修改list。 # Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x#...
Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing(拼接) together the nodes of the first two lists. The sorted list is in ascending order. typedef struct _ListNode { int val; struct _ListNode *next;...
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