#长度长的链表先走m-n步,之后再一次遍历寻找 #方法2: #先走到一个链表的尾部,从尾部开始走; #跳到另一个链表的头部 #如果相遇,则相遇点为重合节点 class Solution(object): def getLinkLenth(self,head): lenth=0 while head!=None: lenth+=1 head=head.next return lenth def getIntersectionNode(self, ...
空间复杂度：O（k），其中k是两个列表中唯一元素的个数。方法5：使用operator.countOf()方法import operator as op# Python code to check if two lists# have any element in common# Initialization of listlist1 = [1, 3, 4, 55]list2 = [90, 1, 22]flag = 0# Using in to check element of#...
If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections. Analysis written by@stellari....
链接:https://leetcode-cn.com/problems/intersection-of-two-linked-lists python # 160.相交链表 # https://leetcode-cn.com/problems/intersection-of-two-linked-lists/solution/intersection-of-two-linked-lists-shuang-zhi-zhen-l/ class ListNode: def __init__(self, val): self.val = val self.ne...
Next } else { curB = headA } } return curA } 题目链接: Intersection of Two Linked Lists : leetcode.com/problems/i 相交链表: leetcode.cn/problems/in LeetCode 日更第 142 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 发布于 2022-06-07 08:18...
Python Intersection and Union Python provides built-in functions to find the intersection and union of two sets or lists. These functions areintersectionandunion. In this article, we will explore these functions and see how they can be used in various scenarios. ...
importoperatorasop# Python code to check if two lists# have any element in common# Initialization of listlist1=[1,3,4,55]list2=[90,1,22]flag=0# Using in to check element of# second list into first listforeleminlist2:ifop.countOf(list1,elem)>0:flag=1# checking conditionifflag==1...
The intersection, union, difference, and symmetric difference of two lists; The sorting method of the list. 1. Delete an element in the list To delete an element in the list, you can usedelandremove. del is to delete elements according to subscripts, and remove is to delete elements accord...
everything = []forchunkinlist_of_lists:everything.extend(chunk) 要比串联方法快: 深色代码主题 复制 everything = []forchunkinlist_of_lists:everything = everything + chunk 排序 你可以用sort函数将一个列表原地排序(不创建新的对象): 深色代码主题 ...
We can further reduce the last two lines of code in the above version of our function to one line by removing the unnecessary use of thefoundvariable. Rather than assigning the results of theintersectionto thefoundvariable and returning that, just return theintersection: ...