[C topic] the penultimate node in the Niuke linked list

Last in the list k Nodes _ Niuke Tiba _ Cattle from  

  Ideas ： Mark two pointers aim and cur from pListhead Start , Let the previous pointer first cur Move in advance k Nodes , Move the two pointers synchronously , Until the front cur Pointer for NULL,aim The position of the pointer is the penultimate k individual .

（IO type OJ topic , It can be improved step by step through the test cases he gives .）

The test case ：

1,{1,2,3,4,5}
5,{1,2,3,4,5}
100,{}
6,{1,2,3,4,5}
0,{1,2,3,4,5}
10,{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}

Expected output ：

{5}
{1,2,3,4,5}
{}
{}
{}
{11,12,13,14,15,16,17,18,19,20}

struct ListNode* FindKthToTail(struct ListNode* pListHead, int k ) 
{
    if(pListHead==NULL||k==0)// Empty linked list or k be equal to 0 Return to the empty linked list 
        return NULL;
    struct ListNode* aim=pListHead;
    struct ListNode* cur=pListHead;
    while(k>0)//k Indicates the number of times the front pointer needs to be moved 
    {
        if(cur==NULL)// once cur Equal to null pointer must terminate loop 
            break;
        cur=cur->next;
        k--;// Move once ,k Minus one .
    }
    if(k>0)//k A positive number means break Out of circulation , It indicates that the number of nodes has exceeded , Returns the entire list .
        return NULL;
    while(cur)// The most common situation , Double pointers move side by side , until cur by NULL.
    {
        cur=cur->next;
        aim=aim->next;
    }
    return aim;
}