Reverse Nodes in k-Group

原题：

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

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

解题：

链表就地反转的变异版本，每隔K个反转一下，不够K个的不反转，注意处理一下特殊的条件即可。可以AC的C++代码如下：

[code]/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverse(ListNode* head){
        if(!head || !head->next){
            return head;
        }
        ListNode *first = head, *second = head->next, *temp;
        while(second){
            temp = second->next;
            second->next = first;
            first = second;
            second = temp;
        }
        head->next = NULL;
        return first;
    }

    ListNode* reverseKG(ListNode* head, int k) {
        int length = 0;
        ListNode *pHead = head, *end;
        while(length < k && pHead){
            end = pHead;
            pHead = pHead->next;
            length ++;
        }
        if(length < k){
            return head;
        }else{
            end->next = NULL;
            ListNode* newHead = reverse(head);
            head->next = reverseKG(pHead, k);
            return newHead;
        } 
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        if(!head || !head->next){
            return head;
        }
        head = reverseKG(head, k);
        return head;
    }
};