25.K个一组翻转链表.md

方法一：模拟

这道题可以进行拆解，拆解成对每个部分进行反转链表，而只需要确定pre、start、end、next这几个位置。

在进行反转时要断开pre到start的指针、end到next的指针，然后对start到end部分进行反转，之后再将pre指向反转后的头节点，start的next指向next，重新建立起连接。

pre移动到start、end也移动到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 dummyNode = new ListNode();
        dummyNode.next = head;
        ListNode start, pre = dummyNode, end = dummyNode;
        while (end.next != null) {
            for (int i = 0; i < k && end != null; i++) end = end.next;
            if (end == null) break;
            start = pre.next;
            pre.next = null;
            ListNode next = end.next;
            end.next = null;
            pre.next = reverse(start);
            start.next = next;
            pre = start;
            end = pre;
        }
        return dummyNode.next;
    }

    public ListNode reverse(ListNode head) {
        ListNode pre = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}