160. Intersection of Two Linked Lists

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.

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.

Credits:Special thanks to @stellari for adding this problem and creating all test cases.

Solution 1

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

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
  var lenA = getLen(headA);
  var lenB = getLen(headB);
  let diff = Math.abs(lenA - lenB);

  if (lenA > lenB) {
    while (diff--) headA = headA.next;
  } else {
    while (diff--) headB = headB.next;
  }

  while (headA !== headB) {
    headA = headA.next;
    headB = headB.next;
  }

  return headA;
};

var getLen = function (head) {
  var len = 0;
  while (head) {
    len++;
    head = head.next;
  }
  return len;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).

Solution 2

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

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
  if (!headA || !headB) return null;  

  var nowA = headA;
  var nowB = headB;

  while (nowA !== nowB) {
    nowA = nowA ? nowA.next : headB;
    nowB = nowB ? nowB.next : headA;
  }

  return nowA;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).