19. Remove Nth Node From End of List

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.

很典型的2 pointer题目。先前进n, 然后当快的pointer到tail时候，慢的正好到要删除的前一个。

public ListNode RemoveNthFromEnd(ListNode head, int n) {
        ListNode slow = head;
        ListNode fast = head;
        ListNode sentinel = new ListNode(-1);
        sentinel.next = head;
        
       
        //first get the n distance between slow and fast, as it always valid, no need for valid check
        for(int i =0;i<n;i++)
        {
            fast = fast.next;
        }
        if(fast == null)//the first nood
        {
            sentinel.next = head.next;
            return sentinel.next;
        }
        
        while(fast.next != null)
        {
            slow = slow.next;
            fast = fast.next;
        }
 
        slow.next = slow.next.next;
        return head;
    }