Remove Nth Node From End of List

问题描述

Given a linked list, remove the n^th 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.

算法

代码一

 1 public ListNode removeNthFromEnd(ListNode head, int n) {
 2         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
 3         ListNode p = head, q;
 4         int index = 1;
 5         while (p.next != null) {
 6             map.put(index, p.val);
 7             index++;
 8             p = p.next;
 9         }
10         map.put(index, p.val);
11         int key = index + 1 - n;
12         int val = map.get(index + 1 - n);
13         p = q = head;
14         for (int i = 1; i < key; i++) {
15             q = p;
16             p = p.next;
17         }
18         if (p == q)
19             head = p.next;
20         else {
21             q.next = p.next;
22         }
23         return head;
24     }

代码二

 1 /**
 2  * 
 3  * @author audr00
 4  * A one pass solution can be done using pointers. Move one pointer fast --> n+1 places forward, 
 5  * to maintain a gap of n between the two pointers and then move both at the same speed. Finally, 
 6  * when the fast pointer reaches the end, the slow pointer will be n+1 places behind - just the right
 7  * spot for it to be able to skip the next node.
 8  * Since the question gives that n is valid, not too many checks have to be put in place. Otherwise, 
 9  * this would be necessary.
10  *
11  */
12     public ListNode removeNthFromEnd(ListNode head, int n) {
13 
14         ListNode start = new ListNode(0);
15         ListNode slow = start, fast = start;
16         slow.next = head;
17 
18         //Move fast in front so that the gap between slow and fast becomes n
19         for(int i=1; i<=n+1; i++)   {
20             fast = fast.next;
21         }
22         //Move fast to the end, maintaining the gap
23         while(fast != null) {
24             slow = slow.next;
25             fast = fast.next;
26         }
27         //Skip the desired node
28         slow.next = slow.next.next;
29         return start.next;
30     }
31     public static void main(String[]args){
32         //new RemoveNthFromEnd1().removeNthFromEnd(head, n)
33     }