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...
## 解法二:递归 Recursion class Solution: def mergeTwoLists(self, head1, head2): ## head1 和 head2 头结点指代两个链表 ## 递归的 base,结束条件 if head1 is None: return head2 elif head2 is None: return head1 elif head1.val < head2.val: ## 把小的值 head1 提取出来,拼接上后面...
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...
21: Merge Two Sorted Lists 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...
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 和88. Merge Sorted Array类似,数据结构不一样,这里是合并链表。
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) { ...
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 1. 2. 题解: ...
Python源码: # Definition for singly-linked list.classListNode:def__init__(self,x):self.val=x self.next=NoneclassSolution:defmergeTwoLists(self,l1:ListNode,l2:ListNode)->ListNode:head=ListNode(0)first=headwhilel1andl2:ifl1.val>l2.val:head.next=l2 ...
用一个大小为K的最小堆(用优先队列+自定义降序实现)(优先队列就是大顶堆,队头元素最大,自定义为降序后,就变成小顶堆,队头元素最小),先把K个链表的头结点放入堆中,每次取堆顶元素,然后将堆顶元素所在链表的下一个结点加入堆中。 代码语言:javascript ...