LeetCode: Partition List Solution

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} x
11 * @return {ListNode}
12 */
13// not fair way
14var partition = function (head, x) {
15 const arr = []
16
17 while (head) {
18 arr.push(head.val)
19 head = head.next
20 }
21
22 const nodes = [...arr.filter(el => el < x), ...arr.filter(el => el >= x)].map(
23 el => new ListNode(el)
24 )
25
26 for (let i = 0; i < nodes.length - 1; i++) {
27 nodes[i].next = nodes[i + 1]
28 }
29
30 return nodes[0] || null
31}
32
33var partition = function (head, x) {
34 let [dummy1, dummy2] = [new ListNode(0), new ListNode(0)]
35 let [newHead1, newHead2] = [dummy1, dummy2]
36 while (head) {
37 if (head.val < x) {
38 dummy1 = dummy1.next = head
39 } else {
40 dummy2 = dummy2.next = head
41 }
42 head = head.next
43 }
44 dummy2.next = null // avoid cyclic
45 dummy1.next = newHead2.next
46 newHead1 = newHead1.next
47 return newHead1
48}

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