25.K个一组翻转链表

思路：

把链表节点按照k个一组分组，可以使用一个指针head依次指向每组的头节点，这个指针每次向前移动k步，直至链表结尾，对于每个分组， 先判断它的长度是否大于等于k，若是，就翻转这部分链表，否则不需要翻转

/*** 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 hair = new ListNode(0);hair.next = head;ListNode pre = hair;while(head != null){ListNode tail = pre;//查看剩余部分长度是否大于等于kfor(int i = 0 ;i<k;i++){tail = tail.next;if(tail == null){return hair.next;}}ListNode next = tail.next;ListNode[] reverse = Reverse(head,tail);head = reverse[0];tail = reverse[1];//把子链表重新接回原链表pre.next = head;tail.next = next;pre = tail;head = tail.next;}return hair.next;}//反转链表，返回子链表的头部与尾部public ListNode[] Reverse(ListNode head,ListNode tail){ListNode prev = tail.next;ListNode p = head;while(prev != tail){ListNode next = p.next;p.next = prev;prev = p;p = next;}return new ListNode[]{tail,head};}
}