题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(head->next == nullptr || head == nullptr || k < 2)
        {
            return head;
        }

        int size = 0;

        ListNode *count = head;

        while(count != nullptr)
        {
            size++;
            count = count->next;
        }

        int group = size / k; 

        ListNode *pre = nullptr;
        ListNode *cur = head;
        ListNode *next = nullptr;
        ListNode *tail = head;
        ListNode *nHead = nullptr;
        ListNode *nTail = nullptr;
        ListNode *nNext = nullptr;

        for(int i=1;i<=k*group;i++)
        {
            tail = tail->next;
        }


        for(int i=1;i<=k*group;i++)
        {
            next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = next;
        }

         nTail = tail;
         for(int i=1;i<=k*group;i++)
        {
            if(i % k == 1)
            {
                nHead = pre;
            }
            nNext = pre->next;
            if(i % k == 0)
            {
                pre->next = nTail;
                nTail = nHead;
            }
            pre = nNext;
        }

        return nTail;
    }
};