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 * rever(struct ListNode * left,struct ListNode * right)
{
    struct ListNode * pre = right;
    struct ListNode * nex = NULL;
    while(left != right)
    {
        nex  =left -> next;
        left -> next = pre;
        pre =left;
        left =nex;
    }
    return pre;
}

struct ListNode* reverseKGroup(struct ListNode* head, int k ) {
    // write code here
    if(!head)
    {
        return NULL;
    }
    int n =0;
    struct ListNode * ptr =head;
    while(ptr)
    {
        ptr =ptr ->next;
        n++;
    }
    if(n < k || k ==1)
    {
        return head;
    }
   
    struct ListNode * node = head;
    for(int i =0 ;i<k;i++)
    {
       node = node -> next;
    }
    struct ListNode * res = rever(head,node);
    head -> next = reverseKGroup(node, k);
    return res;
    
   

}