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最开始的想法是先计算出链表的长度length,然后再从头走 length-n 步即是需要的位置了。
AC代码:
/** * 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) { int length=0; ListNode node=head; while(node!=null){ length++; node=node.next; } for(int i=length-n;i>0;i--){ head=head.next; } return head; } }
然后神奇的快慢指针出场打脸了...
两个指针间距n(慢指针+n=快指针),当快指针到达尾部的时候慢指针所处的位置就是我们需要的啦。
AC代码:
/** * 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) { ListNode fast, slow; fast=slow=head; while(n-->0) fast=fast.next; while(fast!=null){ fast=fast.next; slow=slow.next; } return slow; } }
题目来源: http://www.lintcode.com/zh-cn/problem/nth-to-last-node-in-list/