leetcode 19. Remove Nth Node From End of List

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.

删除倒数第n个节点；

那么设定两个指针，一个fase一个slow，两个相距两个距离。

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {

struct ListNode * f,*s,*t;
f=s=head;

int i;
for(i=0;i<n;i++)
{
f=f->next;
}

if(f==NULL)
{
head=head->next;
free(s);
return head;
}

while(f->next != NULL)
{
f=f->next;
s=s->next;
}

t=s->next;
s->next=t->next;
free(t);
return head;

}