链表系列一＞ K 个一组翻转链表

题目：

链接: link

解析：

代码：

/**
 * 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 {
    /**
    prev: 用来头插
    tmp: 每次头插前记录一下，好进行下一次头插
     */
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int n = 0;
        while(cur != null){
            cur = cur.next;
            n++;
        }
        n = n / k;//统计需要翻转链表几次

        cur = head;
        ListNode newHead = new ListNode(0);
        ListNode prev = newHead;

        for(int i = 0; i < n; i++){
            ListNode tmp = cur;
             for(int j = 0; j < k; j++){
                ListNode next = cur.next;
                cur.next = prev.next;
                prev.next = cur;
                cur = next; 
            }
            prev = tmp;
        }

        //把后面不需要逆序的连接上
        prev.next = cur; 
        return newHead.next;
    }
}