Reference: LeetCode EPI 7.2
Difficulty: Hard

Problem

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.

Note:

  • Only constant extra memory $O(1)$ is allowed.
  • You may not alter the values in the list’s nodes, only nodes itself may be changed.

Example: 

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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

Analysis

Iteration

Calculate how many groups should be reversed. Reverse each group’s nodes.

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//       1   2   3   4   5    k =3
// prev  p   nH  rest
//       2   1   3   4   5
// prev  nH  p   rest
//       2   1   3   4   5
// prev      p   nH  rest
//       3   2   1   4   5
// prev  nH      p   rest
//       3   2   1   4   5
//              prev p
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public ListNode reverseKGroup(ListNode head, int k) {
  if (head == null) {
    return null;
  }
  ListNode dummy = new ListNode(-1);
  dummy.next = head;
  int len = getLength(head);
  // reverse groups
  ListNode prev = dummy;
  ListNode p = head;

  for (int i = 0; i < len / k; ++i) {
    int count = k;
    while (count > 1) { // k = 3, and need to do 2 times
      ListNode newHead = p.next;
      ListNode rest = p.next.next;
      p.next = rest; // set the tail to rest
      newHead.next = prev.next; // not "p"
      prev.next = newHead;
      // no need to update p
      --count;
    }
    prev = p;   // set prev as the new group dummy
    p = p.next; // go to next group
  }
  return dummy.next;
}
  
private int getLength(ListNode head) {
  int len = 0;
  while (head != null) {
    ++len;
    head = head.next;
  }
  return len;
}

Time: $O(N)$
Space: $O(1)$