README.md

160. Intersection of Two Linked Lists

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.

Solutions (Python)

1. Set

# 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
        """
        nodes = set()
        while headA:
            nodes.add(headA)
            headA = headA.next
        while headB:
            if headB in nodes:
                return headB
            headB = headB.next
        return None

2. Change Head to Make Same Lengths

# 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
        """
        cnt = 0
        curr = headA
        while curr:
            cnt += 1
            curr = curr.next

        curr = headB
        while curr:
            cnt -= 1
            curr = curr.next

        for _ in range(cnt):
            headA = headA.next
        for _ in range(-cnt):
            headB = headB.next

        while headA != headB:
            headA = headA.next
            headB = headB.next

        return headA

3. Two Pointers

# 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
        """
        currA = headA
        currB = headB
        while currA != currB:
            currA = currA.next if currA else headB
            currB = currB.next if currB else headA
        return currA