【LeetCode】160. Intersection of Two Linked Lists 解题报告(Python)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/intersection-of-two-linked-lists/description/
题目描述
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
c1 → c2 → c3
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
- If the two linked lists have no intersection at all, return null.
- The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
- Your code should preferably run in O(n) time and use only O(1) memory.
Credits: - Special thanks to @stellari for adding this problem and creating all test cases.
题目大意
找出两个链表的最早公共元素。
解题方法
双指针
第一次遍历时,如果两者的非公共元素的个数正好相等,那么一定能找到相同元素;如果非公共元素个数不等,那么在一次遍历之后,两者的指针的差距就是非公共元素的个数差。这样翻转之后,指针的差距正好弥补了非公共元素的差,这样,第二次遍历要么一定相遇,要么两者没有公共元素,返回None。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
if headA is None or headB is None:
return None
pA = headA
pB = headB
while pA is not pB:
pA = headB if pA is None else pA.next
pB = headA if pB is None else pB.next
return pA
二刷的时候,感觉写的解法更为容易理解。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
len1, len2 = 0, 0
moveA, moveB = headA, headB
while moveA:
len1 += 1
moveA = moveA.next
while moveB:
len2 += 1
moveB = moveB.next
if len1 < len2:
for _ in range(len2 - len1):
headB = headB.next
else:
for _ in range(len1 - len2):
headA = headA.next
while headA and headB and headA != headB:
headA = headA.next
headB = headB.next
return headA
栈
因为后面的元素是相等的,所以使用栈把相等元素都弹出来,那么不等元素就是所求。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
stack1, stack2 = [], []
while headA:
stack1.append(headA)
headA = headA.next
while headB:
stack2.append(headB)
headB = headB.next
pre = None
while stack1 and stack2:
s1 = stack1.pop()
s2 = stack2.pop()
if s1 != s2:
return pre
else:
pre = s1
return pre
日期
2017 年 8 月 27 日
2018 年 11 月 26 日 —— 11月最后一周!
【LeetCode】160. Intersection of Two Linked Lists 解题报告(Python)的更多相关文章
- [LeetCode] 160. Intersection of Two Linked Lists 解题思路
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- LeetCode: Intersection of Two Linked Lists 解题报告
Intersection of Two Linked Lists Write a program to find the node at which the intersection of two s ...
- [LeetCode]160.Intersection of Two Linked Lists(2个链表的公共节点)
Intersection of Two Linked Lists Write a program to find the node at which the intersection of two s ...
- Java for LeetCode 160 Intersection of Two Linked Lists
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- Java [Leetcode 160]Intersection of Two Linked Lists
题目描述: Write a program to find the node at which the intersection of two singly linked lists begins. ...
- LeetCode 160. Intersection of Two Linked Lists (两个链表的交点)
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- [LeetCode] 160. Intersection of Two Linked Lists 求两个链表的交集
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- Leetcode 160. Intersection of two linked lists
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- ✡ leetcode 160. Intersection of Two Linked Lists 求两个链表的起始重复位置 --------- java
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
随机推荐
- DNS域名解析全过程
一张图看懂DNS域名解析全过程 DNS域名解析是互联网上非常重要的一项服务,上网冲浪(还有人在用这个词吗?)伴随着大量DNS服务来支撑,而对于网站运营来说,DNS域名解析的稳定可靠,意味着更多用户 ...
- PPT——一个有情怀的免费PPT模板下载网站!“优品PPT”
http://www.ypppt.com/ PS:再推荐一款免费PPT下载网站 https://www.v5ppt.com/ppt-5-42-1.html
- Excel—分组然后取每组中对应时间列值最大的或者最小的
1.MAX(IF(A:A=D2,B:B)) 输入函数公式后,按Ctrl+Shift+Enter键使函数公式成为数组函数公式. Ctrl+Shift+Enter: 按住Ctrl键不放,继续按Shift键 ...
- 10.Power of Two-Leetcode
Given an integer, write a function to determine if it is a power of two. class Solution { public: bo ...
- 关于写SpringBoot+Mybatisplus+Shiro项目的经验分享一:简单介绍
这次我尝试写一个原创的项目 the_game 框架选择: SpringBoot+Mybatisplus+Shiro 首先是简单的介绍(素材灵感来自英雄联盟) 5个关键的表: admin(管理员): l ...
- 工作学习2-gcc升级引发的崩溃
分享一下调查gcc 8.0下,函数漏写返回值崩溃问题,调查记录. 现在新的硬件,基本操作系统都是redhat 8.0,升级后测试时,发现了一个崩溃问题,记录一下. ================== ...
- 100个Shell脚本——【脚本7】批量建立用户
[脚本7]批量建立用户 编写shell脚本,批量建立用户user_00, user_01, ... user_100并且所有用户同属于users组. 一.脚本 #!/bin/bash group=`c ...
- Output of C++ Program | Set 12
Predict the output of following C++ programs. Question 1 1 #include <iostream> 2 using namespa ...
- When does compiler create default and copy constructors in C++?
In C++, compiler creates a default constructor if we don't define our own constructor (See this). Co ...
- JS21. 使用原生JS封装一个公共的Alert插件(HTML5: Shadow Dom)
效果预览 Shadow DOM Web components 的一个重要属性是封装--可以将标记结构.样式和行为隐藏起来,并与页面上的其他代码相隔离,保证不同的部分不会混在一起,可使代码更加干净.整 ...