# 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.

# Solution

Approach 1: Hash table.

Approach 2: Two pointers: lenA + lenB == lenB + lenA.

# Code (Python)

Approach 2:

    def getIntersectionNode(self, headA, headB):
        if not headA or not headB:
            return None
        node1, node2 = headA, headB
        while True:
            if node1 == node2:
                return node1
            if not node1.next and not node2.next:
                return None
            if node1.next:
                node1 = node1.next
            else:
                node1 = headB
            if node2.next:
                node2 = node2.next
            else:
                node2 = headA

# Code (C++)

Approach 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        unordered_set<ListNode*> visited;
        for (ListNode *p = headA; p != NULL; p = p->next)
        {
            visited.insert(p);
        }
        for (ListNode *p = headB; p != NULL; p = p->next)
        {
            if (visited.find(p) != visited.end())
                return p;
        }
        return NULL;
    }
};

Approach 2:

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *pA = headA;
        ListNode *pB = headB;
        while (pA || pB)
        {
            if (pA == pB)
                return pA;
            if (pA)
                pA = pA->next;
            else
                pA = headB;
            if (pB)
                pB = pB->next;
            else
                pB = headA;
        }
        return NULL;
    }
};