[Algorithm] Find merge point of two linked list

Assume we have two linked list, we want to find a point in each list, from which all the the nodes share the same value in both list. Then we call this point is "merge point".

In the iamge, the merge point shoud be Node(7).

  // Link A length 4, Link B length 5
  // All the node after the merge point should be the same
  // based on that, we can make sure that the merge point should
  // happens in range [x, c] or [y, c], it is not possible to happen at z
  // A      x - x - c - c
  // B  z - y - y - c - c

function Node(val) {
  return {
    val,
    next: null
  };
}

function Link() {
  return {
    head: null,
    tail: null,
    length: ,
    add(val) {
      const newNode = new Node(val);

      if (this.length === ) {
        this.head = newNode;
        this.tail = newNode;
        this.tail.next = null;
      } else {
        let temp = this.tail;
        this.tail = newNode;
        temp.next = this.tail;
      }

      this.length++;
    }
  };
}

 // O ( m + n ), Space: O(1)
function findMergePoint(a, b) {
  // Link A length 4, Link B length 5
  // All the node after the merge point should be the same
  // based on that, we can make sure that the merge point should
  // happens in range [x, c] or [y, c], it is not possible to happen at z
  // A      x - x - c - c
  // B  z - y - y - c - c
  let diff, headA = a.head, headB = b.head;
  if (a.length > b.length) { // O(1)
    diff = a.length - b.length;
    [a, b] = [b, a];
  } else {
    diff = b.length - a.length;
  }

  // reach +diff step for longer
  for (let i = ; i < diff; i++) {
    headB = headB.next; // O(m)
  }

  while (headA != null && headB.val != null ) { // O(n)
    if (headA.val === headB.val) {
      return headA;
    }

    headA = headA.next;
    headB = headB.next;
  } 

  return null;
}

const lA = new Link();
lA.add();
lA.add();
lA.add();
lA.add();

const lB = new Link();
lB.add();
lB.add();
lB.add();
lB.add();
lB.add();

console.log(findMergePoint(lA, lB)); // Object {val: 7, next: Object}