标签:简单 复杂 author 如何 list() 逆序 pac str pop
需要考虑因素,高效应权衡多方面因素01. 先学着实现一个简单的Java版的单项链表
构建任意长度的任意数值的链表, 头插法,顺序遍历输出链表
package com.szs.list;
/**
* 单链表
* @author Administrator
*
*/
public class MyLinkedList {
public int data;
public MyLinkedList next;
public MyLinkedList(int data) {
this.data=data;
this.next=null;
}
public MyLinkedList() {
this.data=-1;
this.next=null;
}
}
02.编写上面的单项链表的逆序输出
高效的输出链表,直接使用栈来存储~~
package com.szs.list;
import java.util.Random;
import java.util.Stack;
public class InverseSingleList {
public static void main(String[] args) {
MyLinkedList head= new MyLinkedList();
createList(head);
inverseList(head);
}
/**
* 构建任意长度的任意数值的链表, 头插法
*/
public static void createList(MyLinkedList head) {
Random random = new Random(System.currentTimeMillis());
int len = random.nextInt(10);
for(int i=0;i<len;i++) {
int data = random.nextInt(100);
MyLinkedList next = new MyLinkedList(data);
next.next = head.next;
head.next = next;
}
/**
* 顺序遍历输出链表
*/
MyLinkedList head2 = head.next;
System.out.println("顺序");
while(head2!=null) {
System.out.print(head2.data+"\t");
head2=head2.next;
}
System.out.println("length="+len);
}
/**
* 高效的输出链表,使用栈来存储
*/
public static void inverseList(MyLinkedList head) {
MyLinkedList head2 = head.next;
Stack<Integer> stack = new Stack<>();
System.out.println("逆序");
while(head2!=null) {
stack.push(head2.data);
head2=head2.next;
}
while(!stack.isEmpty()) {
System.out.print(stack.pop()+"\t");
}
}
}
03.进行测试
顺序
25 69 10 28 23 89 32 2 23 length=9
逆序
23 2 32 89 23 28 10 69 25
-------
顺序
28 35 83 99 88 length=5
逆序
88 99 83 35 28
标签:简单 复杂 author 如何 list() 逆序 pac str pop
原文地址:https://blog.51cto.com/14541438/2438590