Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed. You may not alter the values in the list's nodes, only nodes itself may be changed.

Analyze

初始分析:

每间隔 k 个值进行链表反转, 目前的思考是替换链表的同时维护一个 count, count 达到 k 以后 count 清 0, 同时根据 count 是否达到 k 来判断当前组列表的下一组列表是指向原列表还是反转后的列表, 但是这样的思路需要额外的内存来存储下一组列表。有没有更好的方法呢?

参考网友的 尾插法 思路。

① 将 tail 移到要翻转部分的最后一个元素, 若移动 k 元素之前已到达链表末尾则完成每间隔 k 个值的链表翻转;

② 接着依次把 cur 移到 tail 后面;

该方法的难点一个是如何确定 tail(尾巴) 节点, 另一个是如何穿针引线将 cur 节点移到 tail 节点后面。(这题卡了好久[OMG])

k === 3

prev
tail   head1
dummy    1     2     3     4     5

prev   head1        tail
dummy    1     2     3     4     5
          cur

prev          tail head1
dummy    2     3    1     4     5
          cur

prev    tail       head1
dummy    3     2    1     4     5
            cur

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var reverseKGroup = function(head, k) {
  const dummyHead = new ListNode(0)
  dummyHead.next = head
  let prev = dummyHead
  let tail = dummyHead

  while (true) {
    let count = 0
    while (tail.next && count !== k) {
      tail = tail.next
      count++
    }
    if (count !== k) break

    let head1 = prev.next

    while (prev.next !== tail) {
      let cur = prev.next
      prev.next = cur.next
      cur.next = tail.next
      tail.next = cur
    }

    prev = head1
    tail = head1
  }

  return dummyHead.next
}