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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.
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 public class Solution { 13 /** 14 * @param head: The first node of linked list. 15 * @param n: An integer. 16 * @return: Nth to last node of a singly linked list. 17 */ 18 ListNode nthToLast(ListNode head, int n) { 19 if (head == null) { 20 return null; 21 } 22 ListNode front = head; 23 ListNode node = head; 24 while (n > 0) { 25 if (front == null) { 26 return null; 27 } 28 front = front.next; 29 n--; 30 } 31 while (front != null) { 32 node = node.next; 33 front = front.next; 34 } 35 return node; 36 } 37 }
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原文地址:http://www.cnblogs.com/FLAGyuri/p/5321958.html