intersection-of-two-linked-list.py

# -*- coding: utf-8 -*-
"""
Created on Sat Apr 11 02:23:37 2020

@author: johnoyegbite
"""
# SOLVED!
"""
Problem:
    Write a program to find the node at which the intersection of two singly
    linked lists begins.

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.
"""


def getIntersectionNode(headA, headB):
    """
    :type head1, head1: ListNode
    :rtype: ListNode
    """
    if not headA or not headB:
        return None

    headA_dict = {}
    currA = headA
    while currA:
        headA_dict[currA] = currA
        currA = currA.next

    currB = headB
    while currB:
        if currB in headA_dict:
            return currB
        currB = currB.next

    return None