LeetCode: Remove Linked List Elements Solution

Approach

Keep skipping element by next

For iterative solution, do that first for head to maintain valid head's val

Implementation (iterative)

1/**
2 * Definition for singly-linked list.
3 * function ListNode(val, next) {
4 *     this.val = (val===undefined ? 0 : val)
5 *     this.next = (next===undefined ? null : next)
6 * }
7 */
8/**
9 * @param {ListNode} head
10 * @param {number} val
11 * @return {ListNode}
12 */
13var removeElements = function (head, val) {
14  while (head && head.val === val) {
15    head = head.next
16  }
17
18  let iter = head
19  while (iter && iter.next) {
20    while (iter.next && iter.next.val === val) {
21      iter.next = iter.next.next
22    }
23    iter = iter.next
24  }
25
26  return head
27}

Implementation (recursive)

1var removeElements = function (node, val) {
2  if (!node) return null
3  // console.log('(first)', node.val, !node.next ? 'no next' : node.next.val)
4  node.next = removeElements(node.next, val)
5  // console.log('(second)', node.val, !node.next ? 'no next' : node.next.val)
6  return node.val === val ? node.next : node
7}