LinkedList - 25. Reverse Nodes in k-Group

25. Reverse Nodes in k-Group

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.

思路：

题目意思是每k个节点翻转一次链表，如果节点不足k个，就不翻转，这是一个简单的实现题，运用递归的思想，找到翻转链表的起始点，然后翻转k个节点。

代码：

java：

/**

 * 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 count = 0;
        ListNode cur = head;
        while (cur != null && count != k) {
            cur = cur.next;
            count++;
        }
        
        if (count == k) {
            cur = reverseKGroup(cur, k);
            // resever list
            while (count-- > 0) {
                ListNode tmp = head.next;
                head.next = cur;
                cur = head;
                head = tmp;
            }
            head = cur;
        }
        
        return head;
    }
}