Intersection of Two Linked Lists - LeetCode

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)

• Java Solution:

• /**
   * Definition for singly-linked list.
   * public class ListNode {
   *     int val;
   *     ListNode next;
   *     ListNode(int x) {
   *         val = x;
   *         next = null;
   *     }
   * }
   */
  public class Solution {
      public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
          ListNode p = headA;
          ListNode q = headB;
          int lenA = 0;
          int lenB = 0;
          while(p!=null){
              lenA++;
              p = p.next;
          }
          while(q!=null){
              lenB++;
              q=q.next;
          }
          p = headA;
          q= headB;
          
          if(lenA > lenB){
              int diff = lenA-lenB;
              for(int i=0;i<diff;i++)
                  p = p.next;
          }else if(lenA < lenB){
              int diff = lenB-lenA;
              for(int i=0;i<diff;i++)
                  q = q.next;
          }
          
          while (p!=null && q!=null){
              if(p.val == q.val)
                  return p;
              p = p.next;
              q = q.next;
          }
          return null;
      }
  }