LeetCode: Rotate List Solution

Approach

Find the node for the trim to start, call this the trim list

Attach the tail of the trim list to original head

Detach the node before trim list, point its

to , to avoid cyclic

Implementation

1var rotateRight = function (head, k) {
2  if (!head) return null
3
4  let size = 0
5  for (let iter = head; iter; iter = iter.next) size++
6
7  k %= size
8  k = size - k - 1
9
10  let tail = head
11  let beforeTrimStart = head
12
13  while (tail.next) tail = tail.next
14  while (k--) beforeTrimStart = beforeTrimStart.next
15
16  tail.next = head
17  const trimStart = beforeTrimStart.next
18  beforeTrimStart.next = null
19
20  return trimStart
21}

Implementation (break and rebuild)

1var rotateRight = function (head, k) {
2  if (!head) return null
3
4  const llistToArray = llist => {
5    let iter = llist
6    const res = []
7    while (iter) {
8      res.push(iter.val)
9      iter = iter.next
10    }
11    return res
12  }
13  const arrayToLlist = arr => {
14    let dummy = (iter = new ListNode())
15    arr.forEach(el => {
16      const node = new ListNode(el)
17      iter.next = node
18      iter = iter.next
19    })
20    return dummy.next
21  }
22
23  const arr = llistToArray(head)
24  k %= arr.length
25  k = arr.length - k
26
27  return arrayToLlist([...arr.slice(k), ...arr.slice(0, k)])
28}