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fix comflict
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hejj authored and hejj committed Mar 7, 2017
2 parents f78aaa7 + ba80bf8 commit 58bb4f3
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22 changes: 21 additions & 1 deletion .gitignore
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*.class
# Mobile Tools for Java (J2ME)
.mtj.tmp/

# Package Files #
*.jar
*.war
*.ear

# virtual machine crash logs, see http://www.java.com/en/download/help/error_hotspot.xml
hs_err_pid*

#ide config
.metadata
.recommenders
.idea/
*.iml
rebel.*
.rebel.*

target
11 changes: 11 additions & 0 deletions .project
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<?xml version="1.0" encoding="UTF-8"?>
<projectDescription>
<name>coding2017</name>
<comment></comment>
<projects>
</projects>
<buildSpec>
</buildSpec>
<natures>
</natures>
</projectDescription>
2 changes: 2 additions & 0 deletions .settings/org.eclipse.core.resources.prefs
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eclipse.preferences.version=1
encoding/<project>=UTF-8
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package com.github.Ven13.coding2017.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 originalLen = origin.length;

int len = originalLen;

int temp;

for(int i = 0; i < (originalLen/2); i++){

temp = origin[len - i - 1];

origin[len - i - 1] = origin[i];

origin[i] = temp;

}
}

/**
* 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){

// 定意新数组长度变量
int newLength = 0;
// 获得新数组长度值
for (int i = 0; i < oldArray.length; i++) {

if(oldArray[i] != 0) {

newLength++;

}

}

// 定意新数组
int[] newArray = new int[newLength];
// 定意新数组下标变量
int n = 0;
// 转存数组
for (int i = 0; i < oldArray.length; i++) {

if(oldArray[i] != 0) {

newArray[n] = oldArray[i];// 转存数组
n++;// 新数组下标偏移

}

}

//返回生成的新数组
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){

int[] newArray = new int[array1.length + array2.length];

int k = 0;

String inNum = "";

for(int i = 0; i < array1.length; i++) {

for(int j = 0; j < array2.length; j++) {

if (array1[i] < array2[j]) {
if (inNum.indexOf(array1[i] + "|") < 0) {
newArray[k++] = array1[i];
inNum += array1[i] + "|";
}

} else if (array1[i] == array2[j]) {
if (inNum.indexOf(array1[i] + "|") < 0) {
newArray[k++] = array1[i];
inNum += array1[i] + "|";
}
} else {
if (i == array1.length - 1) {
if (inNum.indexOf(array1[i] + "|") < 0) {
newArray[k++] = array1[i];
inNum += array1[i] + "|";
}
} else {
if (inNum.indexOf(array2[j] + "|") < 0) {
newArray[k++] = array2[j];
inNum += array2[j] + "|";
}
}

}

}

}

return newArray;
}
/**
* 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];
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){
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){

int[] newArray = new int[max];

int k = 0;

boolean isN = true;

if (max > 2) {

for (int i = 2; i < max; i++) {

isN = true;

for (int j = 2; j < max; j++) {

if (i % j == 0 && i != j) {

isN = false;

}
}

if (isN) {

newArray[k++] = i;

}

}

} else if (max == 2) {

newArray[0] = 2;
k++;
} else {

return null;

}

int[] newArray2 = new int[k];

for(int i = 0; i < k; i ++) {

newArray2[i] = newArray[i];

}

return newArray2;
}

/**
* 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 i, j, k;

int sum;

k = 0;

for(i = 1; i <= max; i++) {


sum = 0;

for(j = 1; j < i; j++) {

if(i % j == 0) {

sum += j;

}

}

if(i == sum)
k++;
}

int[] newArray = new int[k];

k = 0;

for(i = 1; i <= max; i++) {


sum = 0;

for(j = 1; j < i; j++) {

if(i % j == 0) {

sum += j;

}

}

if(i == sum) {

newArray[k] = i;

k++;

}

}


return newArray;
}

/**
* 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){

String str = "";

for(int i = 0; i < array.length; i++) {

str += array[i] + seperator;

}

str = str.substring(0, str.length() - 1);

return str;
}



}
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package com.github.Ven13.coding2017.array.test;

import static org.junit.Assert.*;

import org.junit.Test;

import com.github.Ven13.coding2017.array.ArrayUtil;

public class ArrayUtilTest {

@Test
public final void testMerge() {
ArrayUtil arrayUtil = new ArrayUtil();
int[] a1 = {3, 5, 7, 8};
int[] a2 = {4, 5, 6, 7};
int[] a3 = {};
a3 = arrayUtil.merge(a1, a2);
assertEquals(3, a3[0]);
assertEquals(4, a3[1]);
assertEquals(5, a3[2]);
assertEquals(6, a3[3]);
assertEquals(7, a3[4]);
assertEquals(8, a3[5]);

}

@Test
public final void testgetPrimes() {
ArrayUtil arrayUtil = new ArrayUtil();
int max = 23;
int[] a1 = {};
a1 = arrayUtil.getPrimes(max);
assertEquals(3, a1.length);
assertEquals(1, a1[0]);
assertEquals(2, a1[1]);
assertEquals(3, a1[2]);
}

@Test
public final void testgetPerfectNumbers() {
ArrayUtil arrayUtil = new ArrayUtil();
int max = 6;
int[] a1 = {};
a1 = arrayUtil.getPerfectNumbers(max);
}
}
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