LeetCode – Intersection of Two Linked Lists (Java)

Problem

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.

Java Solution

First calculate the length of two lists and find the difference. Then start from the longer list at the diff offset, iterate though 2 lists and find the node.

/**
 * 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) {
        int len1 = 0;
        int len2 = 0;
        ListNode p1=headA, p2=headB;
        if (p1 == null || p2 == null)
            return null;
 
        while(p1 != null){
            len1++;
            p1 = p1.next;
        }
        while(p2 !=null){
            len2++;
            p2 = p2.next;
        }
 
        int diff = 0;
        p1=headA;
        p2=headB;
 
        if(len1 > len2){
            diff = len1-len2;
            int i=0;
            while(i<diff){
                p1 = p1.next;
                i++;
            }
        }else{
            diff = len2-len1;
            int i=0;
            while(i<diff){
                p2 = p2.next;
                i++;
            }
        }
 
        while(p1 != null && p2 != null){
            if(p1.val == p2.val){
                return p1;
            }else{
 
            }
            p1 = p1.next;
            p2 = p2.next;
        }
 
        return null;
    }
}
Category >> Algorithms >> Interview  
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  1. Shawn on 2015-4-29

    I think we should not do if(p1.val == p2.val){, but just if (p1 == p2)..

  2. disqus_NInBSQ2Jb7 on 2015-6-21

    the last while loop is fine ..i dont understand the need to calculate difference and moving pointers

  3. George on 2015-6-24

    Agree!!

  4. Anup on 2015-6-28

    Also assignments:
    p1=headA;
    p2=headB;

    are done twice unnecessarily.

    BTW, thanks for the program, programcreek 🙂

  5. Anand Varma on 2015-10-30

    Reverse both the lists and iterate the reversed lists as long as the nodes are the same. When the nodes differ, the element before that is the intersection point.

  6. Sitian Liu on 2015-12-22

    What if you aren’t allowed to modify the lists and want to keep the space down to O(1)?

  7. Sitian Liu on 2015-12-22

    The last while loop only works if the lists are the same length. The goal is to start at a point on both list where they have even number of elements left

  8. Ayush on 2016-1-22

    if space is not an issue put both in stack ..then pop can compare til both are same.

  9. ramkrishna on 2016-6-24

    p1.val==p2.val is incorrect as doesn’t compare the all the nodes next to it.
    It only compares with two nodes.Replace it with p1==p2
    ex:
    56 79 78 13 97 64 47 20 80 85
    97 6 46 83 54 47 20 81 85
    It will give ans 47 20 80 85
    But correct ans is 85

  10. Udaydeep Thota on 2016-8-30

    // Time Complexity: O(elements in first list + elements in second list)
    // Space Complexity: O(elements in first list)
    public Node intersectionNodeInList(Node head1, Node head2) {

    if(head1==null || head2==null)
    return null;

    HashSet nodeStoreSet = new HashSet();
    Node current1 = head1;
    Node current2 = head2;

    while(current1!=null) {

    nodeStoreSet.add(current1);
    current1=current1.next;

    }

    while(current2!=null) {

    if(nodeStoreSet.contains(current2))
    return current2;
    current2=current2.next;
    }

    return null;

    }

  11. Mayank Mittal on 2017-4-8

    We can also solve it in O(m+n) using hashSet:

    >public ListNode solution2(ListNode list1, ListNode list2) {
    Set set = new HashSet();

    ListNode n1Current = list1;
    while(n1Current != null){
    set.add(n1Current);
    n1Current = n1Current.next;
    }

    ListNode n2Current = list2;

    while(n2Current != null){

    if(set.contains(n2Current)){
    return n2Current;
    }
    n2Current = n2Current.next;
    }

    return null;
    }

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