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.

Solution

Get the length of both linked lists. If the last node does not equal, then there’s no intersection. Then advance the longer list lenA - lenB steps from the head (so that during iteration, both lists reach the end at the same time) and then iterate through the linked list. Return the first match node.

# 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

        lengthA, lengthB = 1, 1
        nodeA = headA
        while nodeA.next is not None:
            nodeA = nodeA.next
            lengthA += 1
        nodeB = headB
        while nodeB.next is not None:
            nodeB = nodeB.next
            lengthB += 1

        if nodeA != nodeB:
            return None
        nodeA, nodeB = headA, headB
        step = 0
        if lengthA > lengthB:
            while step < lengthA - lengthB:
                nodeA = nodeA.next
                step += 1
        elif lengthA < lengthB:
            while step < lengthB - lengthA:
                nodeB = nodeB.next
                step += 1

        while nodeA != nodeB:
            nodeA = nodeA.next
            nodeB = nodeB.next

        return nodeA