p2=p2.next s=Solution()#融合两个链表q=s.mergeTwoLists(l1,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. Example: Input:1->2->4, 1->3->4Output:1->1->2->3->4->4 利用链表的思想,先创建个空链表p,用于存放比较后的结果。然后对传入的两个链表...
Python代码: AI检测代码解析 # Definition for singly-linked list. class ListNode(object): def __init__(self, x): self.val = x self.next = None class Solution(object): def mergeTwoLists(self, l1, l2): """ :type l1: ListNode
public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ListNode *head=new ListNode(0); ListNode *t=head; while(l1!=NULL && l2!=NULL) { if(l1->val<l2->val) { t->next=l1; l1=l1->next; }else{ t->next=l2; l2=l2->next; ...
public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if (l1 == null) return l2; if (l2 == null) return l1; if (l1.val < l2.val) l1.next = mergeTwoLists(l1.next, l2); else l2.next = mergeTwoLists(l1, l2.next); ...
You problem is a pretty basic Python problem: "Iterate trough two lists in parallel. Add the items of both list sequentially to the list of results." Thanks a lot! I can imagine that Python requires a different way of thinking/solving problems. And I already see that some tasks are way...
return self.mergeTwoLists(lists[0], lists[1]) mid = len(lists) // 2 left = self.mergeKLists(lists[:mid]) right = self.mergeKLists(lists[mid:]) return self.mergeTwoLists(left, right) def mergeTwoLists(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNod...
获取元素数大于 1 的列表的中点。使用Python语言时,//执行除法,不带余数。它将除法结果四舍五入到最接近的整数。这也被称为楼层划分。 使用中点作为参考点,将列表拆分为两半。这是分而治之算法范例的分而治之的一面。 Recursion is leveraged at this step to facilitate the division of lists into halved co...
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, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: # 使用一个哨兵...
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...