next = list1; } else { // 如果 list1 没有结点,表明 list2 已遍历完成, // 则将 list2 直接放在 tail 后面 tail.next = list2; } // 返回合并后的链表的头结点 head_pre.next } } 题目链接: Merge Two Sorted Lists : leetcode.com/problems/m 合并两个有序链表: leetcode-cn.com/...
self.next=NoneclassSolution:defmergeTwoLists(self, l1, l2):""":type l1: ListNode :type l2: ListNode :rtype: ListNode""" if l1 is None and l2 isNone:returnNone new_list=ListNode(0) pre=new_listwhile l1 is not None and l2 is notNone:if l1.val pre.next=l1 l1=l1.nextelse: pre...
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: ListNode :rtype: ListNode"""ifnotl1andnotl2:ret...
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...
10, 'b': 8}dict2 = {'d': 6, 'c': 4}# merge the two dictionaries using the Merge() functionmerged_dict = merge(dict1, dict2)# print the merged dictionaryprint(merged_dict)输出{'d': 6, 'b': 8, 'c': 4, 'a': 10}8. 使用reduce()方法from functools import reducedef merge...
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def mergeTwoLists(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode ...
如果不成立,则将链表 l1 和链表 l2 的下一个节点传入递归调用的 mergeTwoLists 方法,并将返回的结果赋值给链表 l2 的下一个节点,并返回链表 l2。 完整代码 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution(object): def mergeKLists(self, lists): """ :type lists: List[ListNode] :...
在Python中,可以使用"+"运算符将多个列表合并为一个列表。具体操作如下: ```python list1 = [1, 2, 3] list2 = [4, 5, 6] list3 = [7, 8...
# method to merge two dictionaries using the dict() constructor with the union operator (|)defmerge(dict1,dict2):# create a new dictionary by merging the items of the two dictionaries using the union operator (|)merged_dict=dict(dict1.items()|dict2.items())# return the merged dictionary...
{ result[k++] = two[j++]; } return result; } /** * @desc 逐个取出1项插入到另外1个已排序数组中去，相当于选择最小项插入到已排序数组中 * 从第1个数组里依次取出项插入到第2个数组合适位置中，这里采用List以便动态调整 */ static List<Integer> mergeSorted2(List<Int...