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.

快指針前移n-1步，如果fast->next==NULL；說明n=鏈表長度，head被刪除；

否則，fast再前移一位，同時slow也前移，當fast到鏈尾時，slow->next就是倒數第n個節點；

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    struct
 
 ListNode* fast = head;
    struct ListNode* slow = head;
    while(--n>0){
        fast = fast->next;
    }
    if(fast->next == NULL){
        return head->next;
    }else{
        fast = fast->next;
    }
    while(fast->next != NULL){
        fast = fast->next;
        slow = slow->next;
    }
    slow
 
->next = slow->next->next;
    return head;
}

leetcode 19. Remove Nth Node From End of List