LeetCode
LeetCode 160 Intersection Of Two Linked Lists - Easy
160. Intersection of Two Linked Lists
160. Intersection Of Two Linked Lists — Easy
Problem
- Intersection of Two Linked Lists
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 Output: Reference of the node with value = 8 Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 Output: Reference of the node with value = 2 Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
Example 3:
Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 Output: null Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values. Explanation: The two lists do not intersect, so return null.
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.
Solution
### Solution 1 : Two Pointers
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param two ListNodes
# @return the intersected ListNode
def getIntersectionNode(self, headA, headB):
if headA is None or headB is None:
return None
pa = headA # 2 pointers
pb = headB
while pa is not pb:
# if either pointer hits the end, switch head and continue the second traversal,
# if not hit the end, just move on to next
pa = headB if pa is None else pa.next
pb = headA if pb is None else pb.next
return pa # only 2 ways to get out of the loop, they meet or the both hit the end=None
# the idea is if you switch head, the possible difference between length would be countered.
# On the second traversal, they either hit or miss.
# if they meet, pa or pb would be the node we are looking for,
# if they didn't meet, they will hit the end at the same iteration, pa == pb == None, return either one of them is the same,None
### Solution 2: Make a cycle, and find intersection like the question of Linked List Cycle II.
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
class Solution:
"""
@param headA: the first list
@param headB: the second list
@return: a ListNode
"""
def getIntersectionNode(self, headA, headB):
# write your code here
if headA == None or headB == None:
return None
connect = headA
while connect.next != None:
connect = connect.next
connect.next = headA
head = headB
slow, fast = head, head.next ### 初始化快慢指针
while fast != slow: ### 直到两指针相遇
if fast is None or fast.next is None:
return None
fast = fast.next.next
slow = slow.next
findIntersection = head
while findIntersection != slow.next: ### 快慢相遇后，看head 和 slow.next的相遇，并各自速度为1步更新，直到相遇时的head就是入口。
findIntersection = findIntersection.next
slow = slow.next
return findIntersection