Merge pull request #23 from nitasty009/master

第二次作业

group05/371492887/task_02/.classpath

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+<?xml version="1.0" encoding="UTF-8"?>
+<classpath>
+	<classpathentry including="**/*.java" kind="src" output="target/classes" path="src">
+		<attributes>
+			<attribute name="optional" value="true"/>
+			<attribute name="maven.pomderived" value="true"/>
+		</attributes>
+	</classpathentry>
+	<classpathentry kind="con" path="org.eclipse.jdt.junit.JUNIT_CONTAINER/4"/>
+	<classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER/org.eclipse.jdt.internal.debug.ui.launcher.StandardVMType/J2SE-1.5">
+		<attributes>
+			<attribute name="maven.pomderived" value="true"/>
+		</attributes>
+	</classpathentry>
+	<classpathentry kind="con" path="org.eclipse.m2e.MAVEN2_CLASSPATH_CONTAINER">
+		<attributes>
+			<attribute name="maven.pomderived" value="true"/>
+		</attributes>
+	</classpathentry>
+	<classpathentry kind="output" path="target/classes"/>
+</classpath>

group05/371492887/task_02/.gitignore

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+/bin/

group05/371492887/task_02/.project

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+<?xml version="1.0" encoding="UTF-8"?>
+<projectDescription>
+	<name>task_02</name>
+	<comment></comment>
+	<projects>
+	</projects>
+	<buildSpec>
+		<buildCommand>
+			<name>org.eclipse.jdt.core.javabuilder</name>
+			<arguments>
+			</arguments>
+		</buildCommand>
+		<buildCommand>
+			<name>org.eclipse.m2e.core.maven2Builder</name>
+			<arguments>
+			</arguments>
+		</buildCommand>
+	</buildSpec>
+	<natures>
+		<nature>org.eclipse.m2e.core.maven2Nature</nature>
+		<nature>org.eclipse.jdt.core.javanature</nature>
+	</natures>
+</projectDescription>

group05/371492887/task_02/pom.xml

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+<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
+	xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
+	<modelVersion>4.0.0</modelVersion>
+	<groupId>java_207</groupId>
+	<artifactId>task_02</artifactId>
+	<version>0.0.1-SNAPSHOT</version>
+
+	<dependencies>
+		<dependency>
+			<groupId>dom4j</groupId>
+			<artifactId>dom4j</artifactId>
+			<version>1.6.1</version>
+		</dependency>
+	</dependencies>
+
+	<build>
+		<sourceDirectory>src</sourceDirectory>
+		<resources>
+			<resource>
+				<directory>src</directory>
+				<excludes>
+					<exclude>**/*.java</exclude>
+				</excludes>
+			</resource>
+		</resources>
+		<plugins>
+			<plugin>
+				<artifactId>maven-compiler-plugin</artifactId>
+				<version>3.1</version>
+				<configuration>
+					<source />
+					<target />
+				</configuration>
+			</plugin>
+		</plugins>
+	</build>
+</project>

group05/371492887/task_02/src/com/coderising/action/LoginAction.java

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+package com.coderising.action;
+
+/**
+ * 这是一个用来展示登录的业务类， 其中的用户名和密码都是硬编码的。
+ * @author liuxin
+ *
+ */
+public class LoginAction{
+    private String name ;
+    private String password;
+    private String message;
+
+    public String getName() {
+        return name;
+    }
+
+    public String getPassword() {
+        return password;
+    }
+
+    public String execute(){
+            if("test".equals(name) && "1234".equals(password)){
+                this.message = "login successful";
+                return "success";
+            }
+            this.message = "login failed,please check your user/pwd";
+            return "fail";
+    }
+
+    public void setName(String name){
+        this.name = name;
+    }
+    public void setPassword(String password){
+        this.password = password;
+    }
+    public String getMessage(){
+        return this.message;
+    }
+}

group05/371492887/task_02/src/com/nitasty/array/ArrayUtil.java

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+package com.nitasty.array;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.Collections;
+import java.util.List;
+
+public class ArrayUtil {
+
+
+	public static void main(String[] args) {
+		ArrayUtil util=new ArrayUtil();
+
+		int[] origin={7, 9 ,10, 30, 3};
+		util.reverseArray(origin);
+		int[] oldArray={1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5};
+		int[] a1={3, 5,7,9,10};
+		int[] a2={4,6,100,111,132}; 
+
+		System.out.println(Arrays.toString(origin));
+		System.out.println(Arrays.toString(util.removeZero(oldArray)));
+		System.out.println(Arrays.toString(util.merge(a1,a2)));
+		System.out.println(Arrays.toString(util.grow(a1,3)));
+		System.out.println(Arrays.toString(util.fibonacci(100)));
+		System.out.println(Arrays.toString(util.getPrimes(100)));
+		System.out.println(Arrays.toString(util.getPerfectNumbers(1000)));
+		System.out.println(util.join(oldArray,"--"));
+	}
+	/**
+	 * 给定一个整形数组a , 对该数组的值进行置换
+		例如： a = [7, 9 , 30, 3]  ,   置换后为 [3, 30, 9,7]
+		如果     a = [7, 9, 30, 3, 4] , 置换后为 [4,3, 30 , 9,7]
+	 * @param origin
+	 * @return
+	 */
+	public void reverseArray(int[] origin){
+		int dataBus;
+		int len=origin.length;
+		for(int i=0;i<len>>1;i++){
+			dataBus=origin[i];
+			origin[i]=origin[len-i-1];
+			origin[len-i-1]=dataBus;
+		}
+	}
+
+	/**
+	 * 现在有如下的一个数组：   int oldArr[]={1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5}   
+	 * 要求将以上数组中值为0的项去掉，将不为0的值存入一个新的数组，生成的新数组为：   
+	 * {1,3,4,5,6,6,5,4,7,6,7,5}  
+	 * @param oldArray
+	 * @return
+	 */
+
+	public int[] removeZero(int[] oldArray){
+		int[] arrayBus=new int[oldArray.length];
+		int count = 0;
+		//利用中间数值来剔除0
+		for (int i = 0; i < oldArray.length; i++) {
+			if(oldArray[i]!=0)
+				arrayBus[count++]=oldArray[i];
+		}
+		//返回新的数值
+		int[] newArray=new int[count];
+		System.arraycopy(arrayBus, 0, newArray, 0, count);
+
+		return newArray;
+	}
+
+	/**
+	 * 给定两个已经排序好的整形数组， a1和a2 ,  创建一个新的数组a3, 使得a3 包含a1和a2 的所有元素， 并且仍然是有序的
+	 * 例如 a1 = [3, 5, 7,8]   a2 = [4, 5, 6,7]    则 a3 为[3,4,5,6,7,8]    , 注意： 已经消除了重复
+	 * @param array1
+	 * @param array2
+	 * @return
+	 */
+	//插入排序挺快的
+	public int[] merge(int[] array1, int[] array2){
+		int len1=array1.length;
+		int len2=array2.length;
+		List<Integer> longList=new ArrayList<Integer>();
+		List<Integer> shortList=new ArrayList<Integer>();
+		//数组转list，有没有简单的方法啊我擦
+		if(len1>len2){
+			for (int i = 0; i < len1; i++) {
+				longList.add(array1[i]);
+			}
+			for (int i = 0; i < len2; i++) {
+				shortList.add(array2[i]);
+			}
+		}else{
+			for (int i = 0; i < len1; i++) {
+				shortList.add(array1[i]);
+			}
+			for (int i = 0; i < len2; i++) {
+				longList.add(array2[i]);
+			}
+		}
+
+		//将短list中的值插入长list中
+		int j=0;
+		for (int i = 0; i < longList.size(); i++) {
+			if(j==shortList.size())
+				continue;
+			if((Integer)shortList.get(j)<(Integer)longList.get(i)){
+				longList.add(i, shortList.get(j));
+				j++;
+			}else if(((Integer)shortList.get(j)).equals((Integer)longList.get(i))){
+				continue;
+			}else{
+				if(i==(longList.size()-1)){
+					longList.add(shortList.get(j));
+					j++;
+				}
+			}
+		}
+
+		//list再转数组···阿西吧
+		int[] intArray=new int[longList.size()];
+
+		for (int i = 0; i < longList.size(); i++) {
+			intArray[i]=longList.get(i);
+		}
+
+		return intArray;
+	}
+
+	/**
+	 * 把一个已经存满数据的数组 oldArray的容量进行扩展， 扩展后的新数据大小为oldArray.length + size
+	 * 注意，老数组的元素在新数组中需要保持
+	 * 例如 oldArray = [2,3,6] , size = 3,则返回的新数组为
+	 * [2,3,6,0,0,0]
+	 * @param oldArray
+	 * @param size
+	 * @return
+	 */
+	public int[] grow(int [] oldArray,  int size){
+		int len=oldArray.length;
+		int[] newArray=new int[len+size];
+
+		System.arraycopy(oldArray, 0, newArray, 0, len);
+
+		for(int i=len;i<len+size;i++){
+			newArray[i]=0;
+		}
+
+		return newArray;
+	}
+
+	/**
+	 * 斐波那契数列为：1，1，2，3，5，8，13，21......  ，给定一个最大值， 返回小于该值的数列
+	 * 例如， max = 15 , 则返回的数组应该为 [1，1，2，3，5，8，13]
+	 * max = 1, 则返回空数组 []
+	 * @param max
+	 * @return
+	 */
+	public int[] fibonacci(int max){
+		if (max==1)
+				return null;
+
+		int a1=1;
+		int a2=1;
+		int count=1;
+		int dataBus=0;
+		while(a2<max){
+			dataBus=a2;
+			a2+=a1;
+			a1=dataBus;
+			count++;
+		}
+		int[] newArray=new int[count];
+		newArray[0]=newArray[1]=1;
+		for (int i = 2; i < count; i++) {
+			newArray[i]=newArray[i-1]+newArray[i-2];
+		}
+		return newArray;
+	}
+
+	/**
+	 * 返回小于给定最大值max的所有素数数组
+	 * 例如max = 23, 返回的数组为[2,3,5,7,11,13,17,19]
+	 * @param max
+	 * @return
+	 */
+	public int[] getPrimes(int max){
+		List<Integer> list=new ArrayList<Integer>();
+		for (int i = 1; i < max; i++) {
+			if(isPrimeNumber(i))
+				list.add(i);
+		}
+
+		int[] intArr=new int[list.size()];
+		for (int i = 0; i < intArr.length; i++) {
+			intArr[i]=list.get(i);
+		}
+		return intArr;
+	}
+
+	private boolean isPrimeNumber(int n)
+	{
+	    if (n==2)
+	    {
+	        return true;
+	    }
+
+	    if (n%2==0)
+	    {
+	        return false;
+	    }
+
+	    int sqrtn=(int)Math.sqrt((double)n);
+	    boolean flag=true;
+
+	    for (int i=3;i<=sqrtn;i+=2)
+	    {
+	        if (n%i==0)
+	        {
+	            flag=false;
+	        }
+	    }
+	    return flag;
+	}
+
+	/**
+	 * 所谓“完数”， 是指这个数恰好等于它的因子之和，例如6=1+2+3
+	 * 给定一个最大值max， 返回一个数组， 数组中是小于max 的所有完数
+	 * @param max
+	 * @return
+	 */
+	public int[] getPerfectNumbers(int max){
+
+		List<Integer> list=new ArrayList<Integer>();
+
+		for (int i = 1; i < max; i++) {
+			int sum=0;
+			for (int j = 1; j < i; j++) {
+				if(i%j==0)
+					sum+=j;
+			}
+			if(sum==i)
+				list.add(i);
+		}
+
+		int[] intArr=new int[list.size()];
+		for (int i = 0; i < intArr.length; i++) {
+			intArr[i]=list.get(i);
+		}
+		return intArr;
+	}
+
+	/**
+	 * 用seperator 把数组 array给连接起来
+	 * 例如array= [3,8,9], seperator = "-"
+	 * 则返回值为"3-8-9"
+	 * @param array
+	 * @param s
+	 * @return
+	 */
+	public String join(int[] array, String seperator){
+		StringBuffer buff=new StringBuffer();
+		for (int i = 0; i < array.length; i++) {
+			buff.append(array[i]);
+			if(i<array.length-1)
+				buff.append(seperator);
+		}
+		return buff.toString();
+	}
+
+
+}