题解 | #链表中的节点每k个一组翻转#

采用快慢指针，快指针与慢指针相距k个节点，每次分组后采用头插法让快指针到慢指针的节点翻转。如果最后不够k个节点，则直接拼接链表。注意每次分组后需要用新链表的尾节点去链表剩下没有遍历的节点以保证链表不会断。这里没有去记录尾节点，而是通过头节点去遍历得到尾节点，应该可以优化直接记录下尾节点。

/**
 * struct ListNode {
 *    int val;
 *    struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        ListNode* new_head=NULL;
        ListNode* fast = head;
        ListNode* slow = head;
        ListNode* group_head = NULL;
        while(fast)
        {
            group_head = NULL;
            //快指针先移动k个位置去占位;
            int group = k;
            while(group>0)
            {
                if(fast == NULL)
                    break;
                fast = fast->next;
                group--;
            }

            if(group==0)
            {
                //够k个位置;
                ListNode* tmp = NULL;
                while(slow && slow!=fast)
                {
                    tmp = slow->next;
                    //采用头插法;
                    if(group_head == NULL)
                    {
                        group_head = slow;
                        slow->next = NULL;
                    }else{
                        slow->next = group_head;
                        group_head = slow;
                    }
                    slow = tmp;
                }

                if(!new_head)
                {
                    new_head = group_head;
                }else
                {
                    ListNode* p = new_head;
                    while(p->next) p = p->next;
                    p->next = group_head;
                }
            }else
            {
                if(!new_head)
                {
                    new_head = slow;
                }else
                {
                    ListNode* p = new_head;
                    while(p->next) p = p->next;
                    p->next = slow;
                }
                break;
            }
        }
        return new_head;
    }
};