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;
        }
    }

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