题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 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
 */
struct ListNode* reverseKGroup(struct ListNode* head, int k ) {
    // write code here
    struct ListNode* pre = NULL;
    struct ListNode* next = NULL;
    struct ListNode* phead = NULL;  //返回指针
    struct ListNode* p = NULL;

    struct ListNode* pconvhead = NULL;  //记录每次翻转的初始节点指针
    struct ListNode* plastconvhead = NULL; 
    int i=0;
    int j=0;
    int length=0;

    //返回的链表头
    phead = head;
    p = head;

    if(head ==NULL||head->next == NULL)
        return head;
    else
     {

        //遍历一遍获得链表长度
        while(p)
        {
            p=p->next;
            length++;
        }
        if(k==1||length/k==0)
            return head;

        plastconvhead = head;

        for(j=0;j<length/k;j++)
        {
            for(i=0;i<k;i++)
            {
                if(i==0)
                    pconvhead = head;
                next = head->next;
                head->next = pre;
                pre = head;
                head = next;
            }

            //记录链表头
            if(j==0)
                phead = pre;

            if(j != 0)
            {
               //上一次反转后的末尾指向下一次翻转完成的起始
                plastconvhead->next = pre;
                plastconvhead = pconvhead;
            }

            pre = NULL;  
            next = NULL;
        }

	    //将最后的部分连接起来
        pconvhead->next = head;

        return phead;
     }
}