每天一道算法题7 查找链表中倒数第k个结点 ； 输入一个单向链表。如果该链表的结点数为奇数，输出中间的结点；如果链表结点数为偶数，输出中间两个结点前面的一个

typedef struct list_node{

int i;

list_node *next;

}LIST_NODE;

LIST_NODE *find_node(LIST_NODE *head, int k)

{

LIST_NODE *p1 = head;

int count = 0;

while(p1 && count++ < k)

p1 = p1->next;

if (!p1)

{

return NULL;

}

LIST_NODE *p2 = head;

while(p1 && p1->next)

{

p1 = p1->next;

p2 = p2->next;

}

return p2;

}

LIST_NODE *find_node(LIST_NODE *head)

{

LIST_NODE *p1, *p2;

if (head == NULL || head->next == NULL)

{

return head;

}

p1 = head;

p2 = head->next;

while(p2->next && p2->next->next)

{

p2 = p2->next->next;

p1 = p1->next;

}

if (p2->next)

{

return p1->next;

}

return p1;

}

int _tmain(int argc, _TCHAR* argv[])

{

LIST_NODE *head = new LIST_NODE;

head->i = 0;

LIST_NODE *temp = head;

for (int j = 0; j < 11; j++)

{

LIST_NODE *p = new LIST_NODE;

p->i = ++j;

temp->next = p;

temp = p;

}

temp->next = NULL;

for (LIST_NODE *temp = head; temp; temp = temp->next)

{

cout << " " << temp->i << " ";

}

LIST_NODE *temp2 = find_node(head);

if(temp2)

cout << "/n" << temp2->i << endl;

system("pause");

return 0;

}