160. Intersection of Two Linked Lists

Description

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.

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.

• Each value on each linked list is in the range [1, 10^9].

• Your code should preferably run in O(n) time and use only O(1) memory.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], 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 [5,6,1,8,4,5]. 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 = [1,9,1,2,4], 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 [1,9,1,2,4]. 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.

Solutions

ListNode

// Definition for singly-linked list.
public class ListNode {
    int val;
    ListNode next;
    
    ListNode(int x) {
        val = x;
        next = null;
    }
}

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA == null || headB == null) {
            return null;
        }
        
        int m = findLength(headA);
        int n = findLength(headB);
        
        if(m > n) {
            headA = nodesToSkip(headA, Math.abs(m-n));
        } else {
            headB = nodesToSkip(headB, Math.abs(m-n));
        }
        
        while(headA != null && headB != null) {
            if(headA == headB) {
                return headA;
            }
            headA = headA.next;
            headB = headB.next;
        }
        
        return null;
    }
    
    private int findLength(ListNode curr) {
        int len = 0;
        while(curr != null) {
            len++;
            curr = curr.next;
        }
        return len;
    }
    
    private ListNode nodesToSkip(ListNode curr, int skip) {
        while(skip-- > 0) {
            curr = curr.next;
        }
        return curr;
    }
}

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