Solution
Approach 1: Brute Force
For each node a_{i} in list A, traverse the entire list B and check if any node in list B coincides with a_{i}.
Complexity Analysis

Time complexity : .

Space complexity : .
Approach 2: Hash Table
Traverse list A and store the address / reference to each node in a hash set. Then check every node b_{i} in list B: if b_{i} appears in the hash set, then b_{i} is the intersection node.
Complexity Analysis

Time complexity : .

Space complexity : or .
Approach 3: Two Pointers
 Maintain two pointers and initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
 When reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when reaches the end of a list, redirect it the head of A.
 If at any point meets , then / is the intersection node.
 To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), would reach the end of the merged list first, because traverses exactly 2 nodes less than does. By redirecting to head A, and to head B, we now ask to travel exactly 2 more nodes than would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
 If two lists have intersection, then their last nodes must be the same one. So when / reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
Complexity Analysis

Time complexity : .

Space complexity : .