LeetCode 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.

解析：
分组翻转链表，根据给定的数值K，按照K分组进行逆序，如果最后剩余元素不足K，则保持原样。
考虑递归的方式处理，先把前K个元素单独拿出来，将这K的元素逆序，然后再与K元素之后的元素进行连接，后面的元素也是同样的处理。具体见代码注释。

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
    public:
    //递归的方式进行链表翻转
    ListNode* reverseByRecursion(ListNode *head) {
        if(head == NULL || head->next == NULL)
            return head;
        ListNode* new_list = reverseByRecursion(head->next);
        head->next->next = head;
        head->next = NULL;
        return new_list;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* temp = head;
        for (int i = 1; i < k && temp != NULL; i++)//先找到前K个元素 {
            temp = temp->next;
        }
        if(temp == NULL)
            return head;
        ListNode *part_head = temp->next;
        //后面元素
        temp->next = NULL;
        //前K的元素与后面元素的连接关系断掉，将前K的元素单独拿出来
        ListNode *new_head = reverseByRecursion(head);
        //前K个元素逆序
        ListNode *new_temp = reverseKGroup(part_head, k);
        //后面元素依次处理
        head->next = new_temp;
        //将翻转后的元素连接起来
        return new_head;
    }
}
;

运行结果：