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.

思路：快慢指针，相距为n;
代码：
/*
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode next;
* };
/
struct ListNode removeNthFromEnd(struct ListNode head, int n) {
if(head==NULL) return NULL;
struct ListNode fast,slow,dummy;
dummy=(struct ListNode)malloc(sizeof(struct ListNode));
dummy->next=head;
slow=fast=dummy;
while(n--){
fast=fast->next;
}
while(fast->next){
fast=fast->next;
slow=slow->next;
}
slow->next=slow->next->next;
return dummy->next;
}