每天一个小算法（5）----找到链表倒数第K个结点

估计这个问题在面试中被问烂了。

思路是先找到正数的第K个结点的指针pT，然后和指向头结点的指针pN一起向后移动，直到第K个指针指向NULL，此时pN指向的结点即倒数第K个结点。

如图：

 #include <stdio.h>
 #include <time.h>
 #include <stdlib.h>
 typedef struct Node
 {
     int data;
     Node* next;
 }Node, *List;
 
 List createList(int num) //随机生成数字，构造链表
 {
     List aList = (List)malloc(sizeof(Node));
     aList->next = NULL;
     aList->data = ;
     Node* qT = aList;
 
      for ( int i=; i< num; ++i)
      {
          Node* pTN = (Node*)malloc(sizeof(Node));
          pTN->data = rand()%;
          pTN->next = NULL;
          qT->next = pTN;
          qT = pTN;
      }
      return aList;
 }
 
 void printList(List aList)    //正序打印链表
 {
     if ( aList == NULL || aList->next == NULL )
         return;
 
     Node* pT = aList->next;
     printf("element of the list:\n\t");
     while( pT != NULL )
     {
         printf("%d ", pT->data);
         pT = pT->next;
     }
 
     printf("\n");
 }
 
 //链表倒数第K个节点主算法
 Node* rfind_K_node(List aList, int K)
 {
     if ( aList == NULL || K <= )
         return NULL;
 
     Node* pT = aList;
     Node* pN = aList;
     for ( int i=; i< K; ++i )
     {
         pT = pT->next;
     }
 
     while ( pT != NULL )
     {
         pN = pN->next;
         pT = pT->next;
     }
 
     return pN;
 }
 
 void deleteList(List aList)    //删除链表
 {}
 
 int main(int argc, char const *argv[])
 {
      srand((int)time());
     List aList = createList();
     printList(aList);
     printf("The bottom %d Node data is %d\n", , rfind_K_node(aList, )->data);
     deleteList(aList);
 
     return ;
 }