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

public ListNode removeNthFromEnd(ListNode head, int n) {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        ListNode p = head, q;
        int index = 1;
        while (p.next != null) {
            map.put(index, p.val);
            index++;
            p = p.next;
        }
        map.put(index, p.val);
        int key = index + 1 - n;
        int val = map.get(index + 1 - n);
        p = q = head;
        for (int i = 1; i < key; i++) {
            q = p;
            p = p.next;
        }
        if (p == q)
            head = p.next;
        else {
            q.next = p.next;
        }
        return head;
    }

代码二

/**
 * 
 * @author audr00
 * A one pass solution can be done using pointers. Move one pointer fast --> n+1 places forward, 
 * to maintain a gap of n between the two pointers and then move both at the same speed. 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.
 *
 */
    public ListNode removeNthFromEnd(ListNode head, int n) {

        ListNode start = new ListNode(0);
        ListNode slow = start, fast = start;
        slow.next = head;

        //Move fast in front so that the gap between slow and fast becomes n
        for(int i=1; i<=n+1; i++)   {
            fast = fast.next;
        }
        //Move fast to the end, maintaining the gap
        while(fast != null) {
            slow = slow.next;
            fast = fast.next;
        }
        //Skip the desired node
        slow.next = slow.next.next;
        return start.next;
    }
    public static void main(String[]args){
        //new RemoveNthFromEnd1().removeNthFromEnd(head, n)
    }