题目：

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.

例子:

Example 1:

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.

问题解析：

给定单向链表，以k个节点作为一组进行链表的翻转。如果剩下的节点不满k个，则不进行翻转。

链接：

• LeetCode：https://leetcode.com/problems/reverse-nodes-in-k-group/description/

思路标签

分步进行：先统计，后翻转

解答：

• 题目其实不难，关键是理清楚翻转的操作，以及所需要的节点；
• 首先统计后面的节点是否满足k个作为一组进行翻转，若不满足则不进行翻转；
• 注意每次都需要记录前一个翻转组的最后一个节点，进而与下一组的翻链表进行连接；
• 注意，返回的链表的头节点是记录第一组翻转后的头节点，与后面的无关，或者如果没有一组进行过翻转，则直接返回原始链表头节点。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        if (k == 1 || head == nullptr)
            return head;

        ListNode *pHead = head;
        ListNode *pNode = head;
        ListNode *pTailPrevious = pNode;
        int first = true;
        while(pNode != nullptr){
            ListNode *pTail = pNode;

            int n = 1;
            // 统计后面是否还存在k个节点
            while(pNode != nullptr && n <= k){
                pNode = pNode->next;
                n++;
            }

            // 如果还存在k个节点则进行翻转并对该k个节点翻转链表与之前的进行连接，否则不进行翻转直接连接。
            if(n > k){
                pNode = pTail;
                ListNode *pPrevious = nullptr;
                n = 1;
                while(n <= k){
                    ListNode *pNext = pNode->next;
                    pNode->next = pPrevious;
                    pPrevious = pNode;
                    pNode = pNext;
                    n++;
                }

                // 判断是否是第一段，第一段的翻转节点为最后的头节点
                if(first == true){
                    pHead = pPrevious;
                    pTailPrevious = pTail;
                    first = false;
                }else{
                    pTailPrevious->next = pPrevious;
                    pTailPrevious = pTail;
                }
            }else{
                // 如果还是第一段则说明不足k个，直接返回链表
                if(first == true){
                    return pHead;
                }
                pTailPrevious->next = pTail;
            }
        }

        return pHead;
    }
};