LeetCode: Reverse Linked List II Solution

Approach

I found it hard to understand one-pass solution, so I go with the so-called cheated solution

• Convert to linked list to array
• Manipulate on that array
• Convert back to linked list

Complexity

Time: O(n)

Space: O(n)

with n is the length of the list

Implementation

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} left
11 * @param {number} right
12 * @return {ListNode}
13 */
14var reverseBetween = function (head, left, right) {
15  left--
16  right--
17  const arr = []
18
19  while (head) {
20    arr.push(new ListNode(head.val))
21    head = head.next
22  }
23
24  const resArr = [
25    ...arr.slice(0, left),
26    ...arr.slice(left, right + 1).reverse(),
27    ...arr.slice(right + 1),
28  ]
29
30  for (let i = 0; i < resArr.length - 1; i++) {
31    resArr[i].next = resArr[i + 1]
32  }
33
34  return resArr[0]
35}