Leetcode 160. Intersection of Two Linked Lists

https://leetcode.com/problems/intersection-of-two-linked-lists/

Easy

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.

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• 鏈表+雙指針。兩種解法：
  • 一種是計算兩個鏈表長度差，讓長一點的鏈表的指針先走，到兩個指針所在鏈表剩餘長度同樣時，再同時日後走。原理是如果兩個鏈表再某一節點相遇，則相遇以後的剩餘鏈表長度必然是同樣的。
  • 另外一種是兩個指針同時日後走，如果一個指針到達尾指針，則指向另外一個鏈表的頭指針，繼續日後走。原理是經過交換頭指針來抵消長度差，如果中途都沒有相遇，則兩個指針都再度到達尾指針，走的長度爲len(linkA) + len(linkB) = len(linkB) + len(linkA)

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution(object):
 8     def getIntersectionNode(self, headA, headB):
 9         """
10         :type head1, head1: ListNode
11         :rtype: ListNode
12         """
13         
14         if headA is None or headB is None:
15             return None
16         
17         pA, pB = headA, headB
18         
19         # if pA meets pB return the common node,
20         # otherwise both reach the tail and return None
21         while pA != pB:
22             # counteract the difference of length with switching head nodes
23             pA = headB if pA is None else pA.next
24             pB = headA if pB is None else pB.next
25             
26         return pA

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