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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• 链表+双指针。两种解法：
  • 一种是计算两个链表长度差，让长一点的链表的指针先走，到两个指针所在链表剩余长度一样时，再同时往后走。原理是若是两个链表再某一节点相遇，则相遇之后的剩余链表长度必然是一样的。
  • 另一种是两个指针同时往后走，若是一个指针到达尾指针，则指向另一个链表的头指针，继续往后走。原理是通过交换头指针来抵消长度差，若是中途都没有相遇，则两个指针都再度到达尾指针，走的长度为len(linkA) + len(linkB) = len(linkB) + len(linkA)

# 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, pB = headA, headB
        
        # if pA meets pB return the common node,
        # otherwise both reach the tail and return None
        while pA != pB:
            # counteract the difference of length with switching head nodes
            pA = headB if pA is None else pA.next
            pB = headA if pB is None else pB.next
            
        return pA