Description:
给你一个链表以及一个k,将这个链表从头指针开始每k个翻转一下。 链表元素个数不是k的倍数，最后剩余的不用翻转。
Explanation:
给出链表 1->2->3->4->5
k = 2, 返回 2->1->4->3->5
k = 3, 返回 3->2->1->4->5
Solution:
首先判断长度能否翻转。如果能翻转，依次将节点插入头部，直至操作k次。然后递归处理。注意每次返回的结果都是k个长度的子串，每次翻转后都需要拼接。
1 -> 2 -> 3 -> 4 -> 5 pre=null;
| |
c cn
1 -> 2 -> 3 -> 4 -> 5 2 -> 1 -> null
| | |
pre c cn
1 -> 2 -> 3 -> 4 -> 5 3 -> 2 -> 1 -> null
| | | |
pre c cn
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { /** * @param head a ListNode * @param k an integer * @return a ListNode */ public ListNode reverseKGroup(ListNode head, int k) { // Write your code here if(k == 1) return head; if(head == null || head.next == null) return head; ListNode current = head; ListNode pre = null; ListNode cnext = null; ListNode res = null; if(len(head , k)){ cnext = current.next; int count = k; while(count >= 1){ count--; cnext = current.next; if(count == 0){ res = current; } current.next = pre; pre = current; current = cnext; } }else{ return head; } ListNode last = res; while(last.next != null){ last = last.next; } ListNode next = reverseKGroup(cnext , k); last.next = next; return res; } public boolean len(ListNode head , int k){ while(head != null){ head = head.next; k--; } return k<=0?true:false; } }