题解 | #链表中的节点每k个一组翻转#

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

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */

    //反转m的下一个到n
    void ReserveList(ListNode* m,ListNode* n)
    {
        ListNode* ss = m->next;
        ListNode* ssnext;
        n = n->next;
        m->next = n;
        //交换到n的下一个
        while(ss != n)
        {
            ssnext = ss->next;
            ss->next = m->next;
            m->next = ss;
            ss = ssnext;
        }
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        if(head == nullptr) return head;
	  //虚拟头结点，方便处理头结点head
        ListNode* virhead = new ListNode(-1);
        virhead->next = head;
        ListNode* fast = virhead;
        ListNode* slow = virhead;
        int pos = 0;
	  //遍历链表，k个反转一次
        while(fast != nullptr)
        {
            if(pos == k)
            {
                ListNode* temp = slow->next;
                ReserveList(slow,fast);
                slow = temp;
                fast = temp;
                pos = 0;
            }
            ++pos;
            fast = fast->next; 
        }
	  //销毁virhead返回他的下一个
        fast = virhead->next;
        delete virhead;
        return fast;
    }
};