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: 

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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.


Solution

Approach #1 (Brute Force) [Time Limit Exceeded]

For each node ai in list A, traverse the entire list B and check if any node in list B coincides with ai.

Complexity Analysis

  • Time complexity : .
  • Space complexity : .

Approach #2 (Hash Table) [Accepted]

Traverse list A and store the address / reference to each node in a hash set. Then check every node bi in list B: if bi appears in the hash set, then bi is the intersection node.

Complexity Analysis

  • Time complexity : .
  • Space complexity : or .

Approach #3 (Two Pointers) [Accepted]

  • Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
  • When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
  • If at any point pA meets pB, then pA/pB 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), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA 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 pA/pB 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 : .

Analysis written by @stellari.

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