//补充一下 链表的逆序操作 添加到listed 类中public Listed reverse() {Listed head = this;Listed p = head;// 单个节点的情况if (p.next == null) {return p;}// 从第二节点的开始Listed q = p;p = p.next;q.next = null;// 防止自循环// 临界条件while (p != null) {Listed temp = q;q = p; //存储p = p.next;q.next = temp;}// 将值付给头节点head = q;return head;}面试中遇到这个问题，写下给大家分享一下
package com.qey.learn;import com.sun.org.apache.xerces.internal.dom.PSVIAttrNSImpl;
import jdk.nashorn.internal.ir.LiteralNode;/*** @ClassName Listed* @Description* @Author qianxl* @Date 2021-03-11 17:12* @Version 1.1**//*** 链表分为包含头节点* 不包含头节点 两种类型*/
public class Listed {private int data;private Listed next;public Listed(int data) {this.data = data;}public Listed getNext() {return next;}public void setNext(Listed next) {this.next = next;}public void showListed() {Listed temp = this.next;if (this == null) {System.out.println("数据为空");}while (temp != null) {System.out.println(temp.data);temp = temp.next;}}public Listed addList(Listed listed) {Listed p = this;if (p.next == null) {p.next = listed;return this;}Listed temp = p.next;while (p != null) {temp = p;p = p.next;}temp.setNext(listed);return this;}/*** 逻辑思路分析：* （1）找到倒数下标p1 比如 倒数2 p1为1 则p2 为2 ;循环 p1=p1.next* 则当p2=p2.next 当p2.next==null为最后一个元素时，则p1就是倒数第二元素* Lastposition =1 时则p1 倒数第二个元素 lastPosition -=1;* @param lastPosition 倒数位置* @return*/public Listed findLastN( int lastPosition) {if(lastPosition<=0){System.out.println("下标从1开始");return null;}lastPosition -=1;//去除head 根节点Listed temp = this.next;Listed p1 = this.next;//临界条件if (p1 == null) {return null;}Listed p2 = p1;while (p2 != null&& lastPosition>0) {lastPosition--;p2 = p2.next;}// 第一种情况就两个节点if (p2.next == null) {this.next = this.next.next;}//第二种情况 多个节点while (p2 != null) {if ( p2.next == null) {temp.next = temp.next.next;//;return this;} else {temp = p1;p1 = p1.next;p2 = p2.next;}}return this;}}class LinkedlistDemo{public static void main(String[] args) {Listed head = new Listed(0);head.addList(new Listed(2));head.addList(new Listed(3));head.addList(new Listed(4));head.addList(new Listed(5));System.out.println(head);Listed lastN = head.findLastN(1);lastN.showListed();}
}
)
