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.

題目要求我們把給出的鏈表按K個一組翻轉，不足K個的不翻轉

這個可以看做是上一個題目兩兩翻轉節點的擴展，我們還是用遞歸來操作，先來確定下大致的處理過程

1.先從鏈表頭開始數k個節點，記錄個數，不滿k個的按k個處理

2.判斷上一步數出的節點個數，小於k則說明不用翻轉，直接返回head就行

3.節點個數等於k，說明需要反轉，利用3指針翻轉鏈表的做法，把這個k節點鏈表翻轉

4.前面幾步把k個節點翻轉了，剩下的節點遞歸調用該方法，然後返回翻轉後鏈表的頭結點

濃縮一下就是，先翻轉k個節點，然後通過遞歸把k個節點之後的鏈表也翻轉

/**
 * 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) {
        ListNode prev = null;
        ListNode cur = head;
        ListNode next = null;
        ListNode check = head;
        int canProceed = 0;
        int count = 0;
        // 檢查鏈表長度是否滿足翻轉
        while (canProceed < k && check != null) {
            check = check.next;
            canProceed++;
        }
        // 滿足條件，進行翻轉
        if (canProceed == k) {
            while (count < k && cur != null) {
                next = cur.next;
                cur.next = prev;
                prev = cur;
                cur = next;
                count++;
            }
            if (next != null) {
                // head 為鏈表翻轉後的尾節點
                head.next = reverseKGroup(next, k);
            }
            // prev 為鏈表翻轉後的頭結點
            return prev;
        } else {
            // 不滿住翻轉條件，直接返回 head 即可
            return head;
        }
    }
}

[LeetCode]25. Reverse Nodes in k-Group k個一組翻轉鏈表