快慢指针-----Nth to Last Node in List

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本文介绍如何在单链表中找到倒数第n个节点，包括算法实现和实例演示。

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Find the nth to last element of a singly linked list.

The minimum number of nodes in list is n.

Have you met this question in a real interview? Yes
Example
Given a List 3->2->1->5->null and n = 2, return node whose value is 1.

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: Nth to last node of a singly linked list. 
     */
    ListNode nthToLast(ListNode head, int n) {
        // write your code here
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode walker = dummy;
        ListNode runner = dummy;
        while (runner.next != null && n>0) {
            runner = runner.next;
            n--;
        }
        while (runner.next != null) {
            runner = runner.next;
            walker = walker.next;
        }
        return walker.next;
    }
}