Intersection of Two Linked Lists

Question

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.

Tags

• Linked List

Thought

The key is the to confirm the difference between the length of the two linked list. Here, we used a pointer (counterPtr in helper function) to represent the length, which means the length of the linked list after this pointer is the difference of the length of two given linked list.

Code

# 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
        """
        def helper(shortList, longList, counterPtr):
            ptrShort, ptrLong = shortList, longList
            while counterPtr:
                counterPtr = counterPtr.next
                ptrLong = ptrLong.next
            while ptrShort != ptrLong:
                ptrShort = ptrShort.next
                ptrLong = ptrLong.next
            return ptrShort

        ptrA, ptrB = headA, headB
        while ptrA and ptrB:
            ptrA = ptrA.next
            ptrB = ptrB.next
        if ptrA is not None:
            ans = helper(headB, headA, ptrA)
        else:
            ans = helper(headA, headB, ptrB)
        return ans