一、题目 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 ret.next=self.mergeTwoLis...
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...
p2=p2.next s=Solution()#融合两个链表q=s.mergeTwoLists(l1,l2)
https://oj.leetcode.com/problems/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. ===Comments by Dabay=== 基本链表的操作。先做一个头节点,用两个指针来记录两个链表的...
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...
题目要求: Merge two sorted linked lists and return it as a new list. The new list should be...
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; ...
Python代码: # 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 :type l2: 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...
Same as the problem - merge two sorted arrays. Linear time complexity. /** * Definition of Interval: * public classs Interval { * int start, end; * Interval(int start, int end) { * this.start = start; * this.end = end; * } * } */ public class Solution { /** * @param l...