以每K个元素为一组逆转单链表

问题：给定一个链表，写一个函数来逆转链表中的每K个节点，K是一个给定的值。

例如：

输入:  1->2->3->4->5->6->7->8->NULL and k = 3 
输出:  3->2->1->6->5->4->8->7->NULL. 

输入:   1->2->3->4->5->6->7->80->NULL and k = 5
输出:  5->4->3->2->1->8->7->6->NULL. 

#include <stdio.h>#include <stdlib.h> /* 链表节点 */typedef struct node{    int data;    struct node* next;}node,*List; /* 以k个元素为一组逆转链表并返回指向新的头结点的指针. */List reverse (List head, int k){    List current = head;    List next;    List prev = NULL;    int count = 0;            /*逆转链表中的前K个节点 */    while (current != NULL && count < k)    {       next  = current->next;       current->next = prev;       prev = current;       current = next;       count++;    }         /* next is now a pointer to (k+1)th node        Recursively call for the list starting from current.       And make rest of the list as next of first node */    if(next !=  NULL)    {  head->next = reverse(next, k); }     /* prev is new head of the input list */    return prev;} /* UTILITY FUNCTIONS *//* Function to push a node */void push(List* head_ref, int new_data){    /* allocate node */    List new_node = (List) malloc(sizeof(node));                 /* put in the data  */    new_node->data  = new_data;     /* link the old list off the new node */    new_node->next = (*head_ref);         /* move the head to point to the new node */    (*head_ref) = new_node;} /* 打印链表节点 */void printList(List node){    while(node != NULL)    {        printf("%d  ", node->data);        node = node->next;    }}     /* 编写程序测试函数 */int main(void){    /* 空链表开始 */    List head = NULL;       /* 创建的链表为： 1->2->3->4->5->6->7->8 */     push(&head, 8);     push(&head, 7);     push(&head, 6);     push(&head, 5);     push(&head, 4);     push(&head, 3);     push(&head, 2);     push(&head, 1);                 printf("Given linked list \n");     printList(head);     head = reverse(head, 3);      printf("\nReversed Linked list \n");     printList(head);      getchar();     return(0);}