需要遍历 list2 中的全部 O(|list2|) 个结点 空间复杂度:O(1) 只需要维护常数个额外变量 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, list...
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...
def mergeTwoList_ext2(list1, list2): if list1 is None: return list2 elif list2 is None: return list1 elif list1.val < list2.val: list1.next = Solution.mergeTwoList_ext2(list1.next, list2) return list1 else: list2.next = Solution.mergeTwoList_ext2(list1, list2.next) return...
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...
最后return的是new_list.next,因为new_list是根节点,new_list.next才是头节点! 二、用递归的解法 classListNode:def __init__(self, x): self.val=x self.next=NoneclassSolution:defmergeTwoLists(self, l1, l2):""":type l1: ListNode :type l2: 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] :...
# method to merge two dictionaries using the dict() constructor with the union operator (|)def merge(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...
解答python:48 ms, 10.8 MB class Solution(object): def mergeTwoLists(self, l1, l2): """ 34620 python技巧 合并两个字典 python 3.5+ 版本 In [1]: a={'x':2,'y':4} In [2]: b={'c':1,'d':3} In [3]: c={'c':3,'y':6} In [4]: w=...'d': 3, 'x': 2, 'y'...
您只看到部分合并列表打印的原因是,您没有打印在最后一个操作中附加(“批量”)的节点。 以下是更正的代码: def merge(self, obj1): # Only need one argument, not two current = self.head # Use self current2 = obj1.head dummy = newnode = node(None) # keep t ...
{ 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...