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Merge pull request #20 from luojunyi/master
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191191717 第二周作业提交
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BlindingDark authored Mar 19, 2017
2 parents a3422a1 + a619dc4 commit 170c8a2
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280 changes: 280 additions & 0 deletions group26/191191717/src/week2/com/coding/basic/ArrayUtil.java
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package week2.com.coding.basic;

import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Set;

public class ArrayUtil
{

/**
* 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[] tempArrays = new int[origin.length];
int j = 0;
for (int i = origin.length - 1; i > -1; i--)
{
tempArrays[j] = origin[i];
j++;
}
}

/**
* 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)
{
// ArrayList不需要知道长度,所以用在此处最方便
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < oldArray.length; i++)
{
if (oldArray[i] != 0)
{
list.add(oldArray[i]);
}
}
int[] newArray = new int[list.size()];
for (int i = 0; i < list.size(); i++)
{
newArray[i] = list.get(i);
}
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
*/

public int[] merge(int[] array1, int[] array2)
{
//Set集合不允许重复值
Set<Integer> set = new HashSet<Integer>();
for (int i : array1)
{
set.add(i);
}
for (int i : array2)
{
set.add(i);
}
int[] newArr = new int[set.size()];
Iterator<Integer> it = set.iterator();
int i = 0;
while (it.hasNext())
{
newArr[i] = it.next();
i++;
}
// 对newArr冒泡排序
for (int j = 0; j < newArr.length - 1; j++)
{
for (int k = 0; k < newArr.length - 1 - j; k++)
{
int temp = 0;
if (newArr[k] > newArr[k + 1])
{
temp = newArr[k];
newArr[k + 1] = temp;
}
}
}
return newArr;
}

/**
* 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[] newArray = Arrays.copyOf(oldArray, oldArray.length + size);
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)
{
// f(n)=f(n-1)+f(n-2)
// f(0)=1 f(1)=1 f(2)=f(0)+f(1)=2
List<Integer> list = new ArrayList<Integer>();
int[] newArr = null;
if (max == 1)
return newArr;
int num = 0;
int x = 1, y = 1;
list.add(1);// f(0)=1;
list.add(1);// f(1)=1;
while (true)
{
num = x + y;
x = y;
y = num;
if (num >= max)
break;
list.add(num);
}
newArr = new int[list.size()];
for (int k = 0; k < list.size(); k++)
{
newArr[k] = list.get(k);
}
return newArr;
}

/**
* 斐波那契数列的递归算法
*
* @param i
* @return
*/
public static int getFiboo(int i)
{
if (i == 1 || i == 2)
return 1;
else
return getFiboo(i - 1) + getFiboo(i - 2);
}

/**
* 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)
{
int[] newArr = null;
List<Integer> list = new ArrayList<Integer>();
for (int i = 1; i < max; i++)
{
boolean flag = isPrime(i);
if (flag)
list.add(i);
}
newArr = new int[list.size()];
for (int k = 0; k < list.size(); k++)
{
newArr[k] = list.get(k);
}
return newArr;
}

public static boolean isPrime(int n)
{
if (n == 1)
return false;
if (n == 2)
return true;
if (n % 2 == 0)
return false;// 偶数肯定不是质数
for (int i = 3; i < n; i += 2)// 去掉偶数的判断
{
if (n % i == 0)
return false;
}
return true;
}

/**
* 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>();
List<Integer> factorList = null;
for (int i = 1; i < max; i++)
{
// 所有的素数不是完数
if (isPrime(i))
{
continue;
}
factorList = getFactor(i);
int count = 0;
for (int j = 0; j < factorList.size(); j++)
{
count += factorList.get(j);
}
if (count != i)
continue;
list.add(i);
}
int[] newArr = new int[list.size()];
for (int k = 0; k < list.size(); k++)
{
newArr[k] = list.get(k);
}
return newArr;
}

/**
* 求一个数的所有因子
**/
public static List<Integer> getFactor(int number)
{
List<Integer> list = new ArrayList<Integer>();
for (int i = 1; i < number; i++)
{
if (number % i != 0)
continue;
list.add(i);
}
return list;
}

/**
* 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 sb = new StringBuffer();
for (int i = 0; i < array.length; i++)
{
if (i == array.length - 1)
{
sb.append(array[i]);
break;
}
sb.append(array[i] + seperator);
}
return sb.toString();
}

}
52 changes: 52 additions & 0 deletions group26/191191717/src/week2/com/coding/litestruts/LoginAction.java
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package week2.com.coding.litestruts;

/**
* 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;
}
}
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