0025.k-个一组翻转链表.java

/*
 * @lc app=leetcode.cn id=25 lang=java
 *
 * [25] K 个一组翻转链表
 *
 * https://leetcode.cn/problems/reverse-nodes-in-k-group/description/
 *
 * algorithms
 * Hard (67.69%)
 * Likes:    1827
 * Dislikes: 0
 * Total Accepted:    406.7K
 * Total Submissions: 600.8K
 * Testcase Example:  '[1,2,3,4,5]\n2'
 *
 * 给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。
 * 
 * k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。
 * 
 * 你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。
 * 
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：head = [1,2,3,4,5], k = 2
 * 输出：[2,1,4,3,5]
 * 
 * 
 * 示例 2：
 * 
 * 
 * 
 * 
 * 输入：head = [1,2,3,4,5], k = 3
 * 输出：[3,2,1,4,5]
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * 链表中的节点数目为 n
 * 1 <= k <= n <= 5000
 * 0 <= Node.val <= 1000
 * 
 * 
 * 
 * 
 * 进阶：你可以设计一个只用 O(1) 额外内存空间的算法解决此问题吗？
 * 
 * 
 * 
 * 
 */

// @lc code=start
/**
 * 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) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy, tail = dummy;
        while (tail != null) {
            for (int i = 0; i < k && tail != null; i++)
                tail = tail.next;
            if (tail == null)
                break;

            ListNode start = pre.next, post = tail.next;
            tail.next = null;
            pre.next = reverse(start);
            start.next = post;
            pre = start;
            tail = pre;
        }

        return dummy.next;
    }

    private ListNode reverse(ListNode head) {
        ListNode current = head;
        head = null;
        while (current != null) {
            ListNode next = current.next;
            current.next = head;
            head = current;
            current = next;
        }

        return head;
    }
}
// @lc code=end