Description
Find the nth to last element of a singly linked list.
The minimum number of nodes in list is n.
Example
Given a List 3->2->1->5->null and n = 2, return node whose value is 1.
解题：给一个链表，求倒数第n个结点的值。先贴一下自己的代码，思路比较简单，遍历两遍，第一遍算结点总数，推出所求结点是第几个，再遍历一次，得出答案。代码如下：
1 /**
2 * Definition for ListNode.
3 * public class ListNode {
4 * int val;
5 * ListNode next;
6 * ListNode(int val) {
7 * this.val = val;
8 * this.next = null;
9 * }
10 * }
11 */
12
13
14 public class Solution {
15 /*
16 * @param head: The first node of linked list.
17 * @param n: An integer
18 * @return: Nth to last node of a singly linked list.
19 */
20 public ListNode nthToLast(ListNode head, int n) {
21 // write your code here
22 int number = 0;
23 int count=0;
24 ListNode p = head;
25 while(p != null){
26 number++;
27 p = p.next;
28 }
29 p=head;
30 while(count != (number-n)){
31 count++;
32 p=p.next;
33 }
34 return p;
35 }
36 }
再贴一下另一中解法，稍微绕个弯子就可以了。定义两个游标，保证前一个与后一个差n，前一个到尾的时候，后一个就是所求值。代码如下：
1 public class Solution {
2 /**
3 * @param head: The first node of linked list.
4 * @param n: An integer.
5 * @return: Nth to last node of a singly linked list.
6 */
7 ListNode nthToLast(ListNode head, int n) {
8 // write your code here
9 ListNode dummy = new ListNode(0);
10 dummy.next = head;
11 ListNode walker = dummy;
12 ListNode runner = dummy;
13 while (runner.next != null && n>0) {
14 runner = runner.next;
15 n--;
16 }
17 while (runner.next != null) {
18 runner = runner.next;
19 walker = walker.next;
20 }
21 return walker.next;
22 }
23 }