0160. Intersection of Two Linked Lists.js

// 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) memory.

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
const getIntersectionNode = function(headA, headB) {
  if (!headA || !headB) {
    return null
  }
  let a = headA
  let b = headB
  while (a !== b) {
    a = a ? a.next : headB
    b = b ? b.next : headA
  }
  return a
}