5. 验证合并结果 我们将合并后的链表打印出来,看看是否满足我们的预期: 1#合并两个有序链表2s =Solution()3newlist_head = s.mergeTwoLists(listnode1.head, listnode2.head)#非递归4#newlist_head = s.mergeTwoListsRecursion(listnode1.head, listnode2.head) # 递归5newlist =ListNode_handle(newlist_hea...
该题目简单,因为已经是两个排序好的链表了。 以下是Python代码,并有测试过程。 #coding:utf-8#Definition for singly-linked list.classListNode(object):def__init__(self, x): self.val=x self.next=NoneclassSolution(object):defmergeTwoLists(self, l1, l2):""":type l1: ListNode :type l2: ListNod...
合并两个有序链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ...
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. Example: Input:1->2->4, 1->3->4Output:1->1->2->3->4->4 ...
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
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...
23. Merge k Sorted Lists Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity. Example: Input:[ 1->4->5, 1->3->4, 2->6 ]Output:1->1->2->3->4->4->5->6 思路: 合并k个有序链表,采用分治的思想,时间复杂度O(nlogk) ...
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...
1 Linked List 和 Python 基础操作 1.1 链表原理 数组之后,链表是第二种基础的数据存储结构。和数组的连续存储不同,链表是的存储方式更加灵活,可以连续也可以不连续。不连续的存储单位通过上一个元素的”next“指针指出,也就是说,单个存储单位不仅存储元素的值,还存储下一个单位的地址信息。
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...