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Merge pull request #17 from skomefen/master
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day3-19
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Mr-ChangK authored Mar 20, 2017
2 parents 115a519 + bd88b50 commit 358e1ce
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package com.coderising.array;

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[] reverse = new int[origin.length];
for(int i=0;i<reverse.length;i++){
reverse[i]=origin[origin.length-i];
}
for(int i=0;i<reverse.length;i++){
reverse[i]=origin[i];
}
}

/**
* 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 rz[]=null;
int index = 0;
for(int i=0;i<oldArray.length;i++){
if(oldArray[i]!=0){
index++;
}
}
rz = new int[index];
index=0;
for(int i=0;i<oldArray.length;i++){
if(oldArray[i]!=0){
rz[index]=oldArray[i];
index++;
}
}
return rz;
}

/**
* 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){
return null;
}
/**
* 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){
return null;
}

/**
* 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){
return null;
}

/**
* 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){
return null;
}

/**
* 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){
return null;
}

/**
* 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){
return null;
}


}
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package com.coderising.litestruts;




import java.util.HashMap;
import java.util.Map;

import org.junit.Assert;
import org.junit.Test;









public class StrutsTest {

@Test
public void testLoginActionSuccess() {

String actionName = "login";

Map<String,String> params = new HashMap<String,String>();
params.put("name","test");
params.put("password","1234");



View view = Struts.runAction(actionName,params);


Assert.assertEquals("/jsp/homepage.jsp", view.getJsp());
Assert.assertEquals("login successful", view.getParameters().get("message"));
}

@Test
public void testLoginActionFailed() {
String actionName = "login";
Map<String,String> params = new HashMap<String,String>();
params.put("name","test");
params.put("password","123456"); //密码和预设的不一�?


View view = Struts.runAction(actionName,params);


Assert.assertEquals("/jsp/showLogin.jsp", view.getJsp());
Assert.assertEquals("login failed,please check your user/pwd", view.getParameters().get("message"));
}
}

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