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Merge pull request #2 from nusubmarine01/master
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merge
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dongyuanlongwang authored Mar 11, 2017
2 parents 8c6675c + c779489 commit d72cd0a
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203 changes: 203 additions & 0 deletions group03/1196051822/3月05日作业的代码/src/ArrayUtil.java
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import java.util.Arrays;

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 tmp ;
int length = origin.length;
for (int i = 0 ; i < length / 2 ; i++) {
tmp = origin[i];
origin[i] = origin[length - i - 1];
origin[length - i - 1] = tmp;
}
}

/**
* 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 size = oldArray.length;
int[] newArray = new int[size];
int repeatTime = 0;
int count = 0;
for (int i = 0 ; i < oldArray.length;i++) {
if (oldArray[i] != 0) {
newArray[count++] = oldArray[i];
}else{
repeatTime++;
}
}
return Arrays.copyOf(newArray,newArray.length - repeatTime);
}

/**
* 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){
int[] newArray = new int[array1.length + array2.length];
int count = 0;
int cursor = 0;
int last = 0;
int repeatTime = 0;
for (int i = 0 ; i < array1.length;i++) {
last = i;
int value1 = array1[i];
if (value1 < array2[cursor]) {
newArray[count++] = value1;
}else{
newArray[count++] = array2[cursor];
if (value1 != array2[cursor]) {
i--;
}else{
repeatTime++;
}
cursor++;
if (cursor == array2.length) {
break;
}
}
}
for (int i = cursor ; i < array2.length ;i++) {
if (newArray[count - 1] == array2[i]) {
continue;
}
newArray[count++] = array2[i];
}

for (int i = last;i < array1.length;i++) {
if (newArray[count - 1] == array1[i]) {
continue;
}
newArray[count++] = array1[i];
}
return Arrays.copyOf(newArray,newArray.length - repeatTime);
}
/**
* 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 = new int[oldArray.length + size];
System.arraycopy(oldArray,0,newArray,0,oldArray.length);
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 new int[]{};
}
int num1 = 1;
int num2 = 1;
int next = 2;
int [] array = new int[max + 1];
array[0] = num1;
int count = 1;
while (next <= max) {
array[count++] = num2;
next = num1 + num2;
num1 = num2;
num2 = next;
}
return Arrays.copyOf(array,count);
}

/**
* 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[] primes = new int[max];
int count = 0;
boolean isPrime = true;
for (int i = 2 ; i < max ;i ++) {
for (int j = 2 ; j < Math.sqrt(i);j++) {
if (i % j == 0){
isPrime = false;
break;
}
}

if (isPrime) {
primes[count++] = i;
}

isPrime = true;
}
return Arrays.copyOf(primes,count);
}

/**
* 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){
int[] array = new int[max];
int count = 0;
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) {
array[count++] = i;
}
}
return Arrays.copyOf(array, count);
}

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


}
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/**
* 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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