25. Reverse Nodes in k-Group

Description

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

You may not alter the values in the list's nodes, only nodes themselves may be changed.

 

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

 

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

 

Follow-up: Can you solve the problem in O(1) extra memory space?

 

Solutions

Solution: Iteration

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
const reverseKGroup = function (head, k) {
  const groups = [];
  const reverseNodes = (node, nextNode) => {
    const nodes = [];

    for (let kth = 1; kth <= k; kth++) {
      nodes.push(node);
      node = node.next;
    }
    nodes[0].next = nextNode;
    let previous = nodes[0];

    for (let index = 1; index < nodes.length; index++) {
      nodes[index].next = previous;
      previous = nodes[index];
    }
    return previous;
  };
  let current = head;
  let kth = 0;

  while (current) {
    kth += 1;
    if (kth === 1) groups.push(current);
    if (kth === k) kth = 0;
    current = current.next;
  }
  let nextNode = kth && kth < k ? groups.pop() : null;

  for (let index = groups.length - 1; index >= 0; index--) {
    nextNode = reverseNodes(groups[index], nextNode);
  }
  return nextNode;
};