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.
分析:
要找到nth to last element,我们需要两个指针,第一个指针先走n步,然后两个指针同时走,知道第一个指针为null.
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 if (head == null || n <= 0) return null; 9 10 ListNode first = head; 11 ListNode last = head; 12 13 for (int count = 1; count <= n; count++;) { 14 first = first.next; 15 } 16 17 while(first != null) { 18 first = first.next; 19 last = last.next; 20 } 21 return last; 22 } 23 }
转载请注明出处:cnblogs.com/beiyeqingteng/
时间: 2024-10-23 07:45:46