链表实现ADT

invoked — Sat, 26 Sep 2009 06:38:00 GMT

import java.util.ArrayList;
import java.util.List;

  /** *//**
   * 链表实现ADT
   * @author BruceLeey
   */
  class Node {

     Object obj;  //数值域
     Node next; //链域

     public Node() {
     }

     public Node(Object value) {

         this.obj = value;
         next = null;
     }
}

class LinkList {

     private Node first;  //头节点，不记录在链表之内
     private int size;  //记录链表节点长度

     public LinkList() {

         first = null;
         size = 0;

     }

     /** *//**
      * 添加节点
      *
      * @param value
      */
     public void addNode(Object value) {
         System.out.println("\n-------------------------添加节点 " + value + " -------------------------");
        Node currentNode = new Node(value);
        currentNode.next = first; //当前节点链域指向头节点
         first = currentNode;        //头节点记录当前节点地址
         size++;
     }

     /** *//**
      * 验证是否为空
      * @return
      */
     public boolean isEmpty() {
         return size == 0;
     }

     /** *//**
      * 删除表头
     * @param value
     */
     public Node removeFirstNode() {
         System.out.println("\n-------------------------移除头节点-------------------------");
         Node temp = first;
         first = first.next;   //指向下一节点
         size--;
         System.out.println("\n移除的表头数据为: " + temp.obj);
         return temp;   //返回删除的节点
     }

     /** *//**
      * 封装长度
      * @return
      */
     public int getSize() {

        return size;
     }

     /** *//**
     * 找出索引之前的节点
      * @param index
      * @return
      */
     public List<Node> getNodeByIndex(int index) {
         System.out.println("\n-------------------------查找" + index + "之前的所有节点-------------------------");
         List<Node> list = new ArrayList<Node>();
         assert (!(index > getSize() - 1 || index < 0));
         Node current = first;   //定位到头节点
         for (int i = 0; i < index; i++) {
             list.add(current);
             current = current.next;   //以此往下移
       }
         for (int j = 0; j < list.size(); j++) {
            System.out.println("\n查找到的数据为:  " + list.get(j).obj);
         }
         return list;
     }

     /** *//**
     * 输出链表
     */
    public void displayNode() {
        System.out.println("\n-------------------------开始输出链表-------------------------");
        assert (!this.isEmpty());
        Node current = first;
        for (int i = 0; i < getSize(); i++) {

            System.out.println("节点为: " + current.obj.toString());
            current = current.next;
        }

    }
}

public class TestAdt {

    public static void main(String[] args) {
        LinkList link = new LinkList();
        for (int i = 0; i < 10; i++) {
            link.addNode("我是节点 " + i);
        }
        link.displayNode();
        Node node = link.removeFirstNode();
        link.displayNode();
        link.getNodeByIndex(5);
        link.displayNode();

    }
}

invoked 2009-09-26 14:38 发表评论

海盗算法

invoked — Sat, 26 Sep 2009 06:35:00 GMT

   问题：
            有17个人围成一圈(编号0~16),从第0号的人开始从1报数，凡报到3的倍数的人离开圈子，
            然后再数下去，直到最后只剩下一个人为止，问此人原来的位置是多少号?

int [] people = new int[17];
    int lastPeople = 0;
    public void getTheLastPeople(){
        for(int i = 0; i < people.length; i++){
            people[i] = i+1;
        }
        int count = 0;
        int countLast = 0;
        int j = 0;
        while(true){
            for(j = 0; j < people.length; j++){
                if(people[j] != 0){
                    count++;
                    people[j] = count;
                    System.out.println("people[" + j + "] = " + people[j]);
                    if (people[j] % 3 == 0) {
                        people[j] = 0;
                        countLast++;
                        if(countLast == 17){
                            lastPeople = j;
                            return;
                        }
                    }
                }
            }
        }
    }

invoked 2009-09-26 14:35 发表评论

BlogJava-bruceleey's columns -随笔分类-算法/数据结构

链表实现ADT

海盗算法