166. Nth to Last Node in List

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：

 /**
  * 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.
      */
     public ListNode nthToLast(ListNode head, int n) {
         // write your code here
         int number = 0;
         int count=0;
         ListNode p = head;
         while(p != null){
             number++;
             p = p.next;
         }
         p=head;
         while(count != (number-n)){
             count++;
             p=p.next;
         }
         return p;
     }
 }

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：

  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;
      }
  }