LeetCode: Add Two Numbers Solution

Dummy head technique in linked list

Use a dummy node to refer back to head

We could also understand this dummy node as a previous node of head

1let dummy = (iter = new ListNode())
2
3// begin iteration
4iter.next = new ListNode(/* calculation for new node */)
5iter = iter.next // new ref for pointer
6// end iteration
7
8return dummy.next

Approach

Use dummy head for later head reference

Iteration will finish when both pointers for 2 lists are null

The rest is handle sum and conditioning when

• sum is larger or equal than 10
• two lists are different in length

Remember to check the remainder at last

Implementation

1var addTwoNumbers = function (l1, l2) {
2  let iter1 = l1
3  let iter2 = l2
4  let dummy = (iter = new ListNode())
5  let remainder = 0
6
7  while (iter1 || iter2) {
8    let val1 = iter1 ? iter1.val : 0
9    let val2 = iter2 ? iter2.val : 0
10    let val = val1 + val2 + remainder
11    remainder = 0
12    if (val >= 10) {
13      val -= 10
14      remainder = 1
15    }
16    iter.next = new ListNode(val)
17    iter = iter.next
18    iter1 = iter1 ? iter1.next : iter1
19    iter2 = iter2 ? iter2.next : iter2
20  }
21  if (remainder === 1) {
22    iter.next = new ListNode(1)
23  }
24
25  return dummy.next
26}