leetcode/160_Intersection_of_Two_Linked_Lists.py at master · annabechang/leetcode · GitHub

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"""

160. Intersection of Two Linked Lists

Easy

2072

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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.

"""
"""
approach:
a = [0,9,1,2,4]
b = [3,2,4]
ab = [0,9,1,2,4,3,2,4]
ba = [3,2,4,0,9,1,2,4]
the overlapped part is at the end

"""

# 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 not headA or not headB:
            return None

        a, b = headA, headB

        while a != b:
            if not a:
                a = headB
            else:
                a = a.next

            if not b:
                b = headA
            else:
                b = b.next

        return a
"""
Success
Details
Runtime: 196 ms, faster than 91.70% of Python online submissions for Intersection of Two Linked Lists.
Memory Usage: 41.9 MB, less than 33.90% of Python online submissions for Intersection of Two Linked Lists.

"""