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.

** 解题报告 ** 使用快慢指针来做。 Using fast and slow pointers.
(1) Move one pointer fast –> n+1 places forward, to maintain a gap of n between the two pointers (2) and then move both at the same speed. (3) Finally, when the fast pointer reaches the end, the slow pointer will be n+1 places behind - just the right spot for it to be able to skip the next node.

Since the question gives that n is valid, not too many checks have to be put in place. Otherwise, this would be necessary.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

    public ListNode removeNthFromEnd(ListNode head, int n) {
      ListNode start = new ListNode(0);
      start.next = head;
      ListNode fast = start;
      ListNode slow = start;
      
      // Move fast in front so that the gap between fast and slow becomes n;
      for(int i = 1; i <= n + 1; i++) {
          fast = fast.next;
      }
      // Moving fast to the end, maintaining the gap
      while(fast != null) {
          fast = fast.next;
          slow = slow.next;
      }
      
      // Skip the desired node
      slow.next = slow.next.next;
      return start.next;
    }