25. Reverse Nodes in k-Group

难度: Hard

题目描述

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?

解题思路

代码实现

解决方案

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null) {
            return null;
        }
        ListNode tailP = head;
        for (int i = 1; i < k; i++) {
            if (tailP != null) {
                tailP = tailP.next;
            }
        }
        if (tailP == null) {
            return head;
        }
        ListNode next = tailP.next;
        tailP.next = null;
        ListNode[] reverseArray = reverseListNode(head, tailP);
        ListNode resHead = reverseArray[0];
        ListNode resTail = reverseArray[1];
        if (resHead == null || resTail == null) {
            return resHead;
        }

        resTail.next = reverseKGroup(next, k);

        return resHead;

    }

    public ListNode[] reverseListNode(ListNode head, ListNode tail) {
        if (head == null || tail == null) {
            return new ListNode[] { head, tail };
        }
        ListNode headP = head;
        ListNode resHead = null;
        ListNode resTail = null;
        while (headP != null) {
            ListNode next = headP.next;
            headP.next = null;
            if (resHead == null) {
                resHead = resTail = headP;
            } else {
                headP.next = resHead;
                resHead = headP;
            }
            headP = next;
        }

        return new ListNode[] { resHead, resTail };
    }

}