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 = 0;
    Node* qT = aList;

     for ( int i=0; i< num; ++i)
     {
         Node* pTN = (Node*)malloc(sizeof(Node));
         pTN->data = rand()%100;
         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:
	");
    while( pT != NULL )
    {
        printf("%d ", pT->data);
        pT = pT->next;
    }

    printf("
");
}

//链表倒数第K个节点主算法
Node* rfind_K_node(List aList, int K)
{
    if ( aList == NULL || K <=0 )
        return NULL;

    Node* pT = aList;
    Node* pN = aList;
    for ( int i=0; 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(0));
    List aList = createList(7);
    printList(aList);
    printf("The bottom %d Node data is %d
", 5, rfind_K_node(aList, 5)->data);
    deleteList(aList);
    
    return 0;
}