Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

 

 

M1: recursive

cur指向head，用cnt计数，先走k步。当走了k步时（即cnt == k），递归，然后反转这一部分的链表

time: O(n), space: O(n)

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode reverseKGroup(ListNode head, int k) {        if(head == null || head.next == null) {            return head;        }        int cnt = 0;        ListNode cur = head;        while(cur != null && cnt < k) {            cur = cur.next;            cnt++;        }                if(cnt == k) {            cur = reverseKGroup(cur, k);            while(cnt-- > 0) {                ListNode next = head.next;                head.next = cur;                cur = head;                head = next;            }            head = cur;        }        return head;    }}