//補(bǔ)充一下 鏈表的逆序操作 添加到listed 類中public Listed reverse() {Listed head = this;Listed p = head;// 單個節(jié)點的情況if (p.next == null) {return p;}// 從第二節(jié)點的開始Listed q = p;p = p.next;q.next = null;// 防止自循環(huán)// 臨界條件while (p != null) {Listed temp = q;q = p; //存儲p = p.next;q.next = temp;}// 將值付給頭節(jié)點head = q;return head;}

面試中遇到這個問題，寫下給大家分享一下?

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**//*** 鏈表分為包含頭節(jié)點* 不包含頭節(jié)點 兩種類型*/ 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("數(shù)據(jù)為空");}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;}/*** 邏輯思路分析：* （1）找到倒數(shù)下標(biāo)p1 比如 倒數(shù)2 p1為1 則p2 為2 ;循環(huán) p1=p1.next* 則當(dāng)p2=p2.next 當(dāng)p2.next==null為最后一個元素時，則p1就是倒數(shù)第二元素* Lastposition =1 時則p1 倒數(shù)第二個元素 lastPosition -=1;* @param lastPosition 倒數(shù)位置* @return*/public Listed findLastN( int lastPosition) {if(lastPosition<=0){System.out.println("下標(biāo)從1開始");return null;}lastPosition -=1;//去除head 根節(jié)點Listed temp = this.next;Listed p1 = this.next;//臨界條件if (p1 == null) {return null;}Listed p2 = p1;while (p2 != null&& lastPosition>0) {lastPosition--;p2 = p2.next;}// 第一種情況就兩個節(jié)點if (p2.next == null) {this.next = this.next.next;}//第二種情況 多個節(jié)點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();} }

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