寻找链表的倒数第N个节点

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package 链表中;public class NthNodeFromEndOfList {/** * 获取链表的长度 */public static int lengthOfList(ListNode head) {int m = 0;ListNode p = head;while(p != null) {m++;p = p.next;}return m;}/* * 方法一：遍历两遍链表 */public static ListNode find01(ListNode head,int n) {ListNode p = head;int m = lengthOfList(head);for (int i = 0; i < m-n; i++) {p = p.next;}return p;}/** * 方法二：只遍历一遍链表，允许多指针 * 思路：定义指针p1,p2 *    指针p2往后移动N位 *   同时将p1,p2往后移动，直到p2遇到null * @param args */public static ListNode find02(ListNode head, int n) {ListNode p1 = head;ListNode p2 = head;for (int i = 0; i < n; i++) {p2 = p2.next;}while (p2 != null) {p2 = p2.next;p1 = p1.next;}return p1;}public static void main(String[] args) {int[] array = {1,2,3,4,5,6};ListNode head01 = ListNode.arrayToList(array);System.out.print("单指针，多遍历： ");System.out.println(find01(head01, 1));ListNode head02 = ListNode.arrayToList(array);System.out.print("多指针，单遍历： ");System.out.println(find02(head02, 3));}}

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