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
     */
    void reverse(ListNode *&left_pre,ListNode *&left,ListNode* &right,ListNode*& right_last)
    {
        ListNode* p=left,* pre=NULL;
        while(p!=right_last)
        {
            ListNode* t=p->next;
            p->next=pre;
            if(p!=right) pre=p;
            p=t;
        }
        left_pre->next=right;
        left->next=right_last;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        if(k==1||head==NULL||head->next==NULL) return head;

        ListNode* H=new ListNode(0);   
        H->next=head;//头节点
        ListNode* pp=head,*pp_pre=H;//遍历
        ListNode* left=head,*left_pre=H,*right,*right_last;
        bool flag_left=0,flag_right=0;
        int i;
        for(i=1;pp;i++)
        {
            if(i%k==1) 
            {
                left_pre=pp_pre;
                left=pp;
                flag_left=1;
            }
            if(i%k==0)
            {
                right=pp;
                right_last=pp->next;
                flag_right=1;
            }
            if(flag_right&&flag_left)
            {
                cout<<i<<" ";
                reverse(left_pre,left,right,right_last);
                flag_left=flag_right=0;
                pp_pre=left;
                pp=right_last;
            }
            else
            {
                pp_pre=pp;
                pp=pp->next;
            }
        }
        return H->next;
    }
};