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
* @return
*/
public int[] reverseArray(int[] origin){
int[] newOrigin = new int[origin.length];
for (int i = 0; i < newOrigin.length; i++) {
newOrigin[i] = origin[origin.length-1-i];
}
return newOrigin;
}
/**
* 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 count = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] == 0) {
count++;
}
}
int newArray[] = new int[oldArray.length-count];
int index = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArray[index] = oldArray[i];
index++;
}
}
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 length = array1.length + array2.length;
int TEMP[]= new int[length];
System.arraycopy(array1, 0, TEMP, 0, array1.length);
System.arraycopy(array2, 0, TEMP, array1.length, array2.length);
for (int i = 0; i < array1.length; i++) {
for (int j = 0; j < array2.length; j++) {
if (array1[i]==array2[j]) {
length--;
}
}
}
for (int i = 0; i < TEMP.length; i++) {
System.out.println(TEMP[i]);
}
int head = 0;
for (int i = 0; i < TEMP.length -1;i++) {
for (int j = i+1; j < TEMP.length; j++) {
if(TEMP[i]>TEMP[j]){
int tempE=TEMP[i];
TEMP[i] = TEMP[j];
TEMP[j] = tempE;
}
}
}
int NewArray[] = new int[length];
int num = 0;
for (int i = 0; i < TEMP.length;i++) {
if (i<TEMP.length-1 && TEMP[i]!=TEMP[i+1] ) {
NewArray[num]=TEMP[i];
num++;
}else if(i==TEMP.length - 1){
NewArray[num]=TEMP[i];
num++;
}
}
for (int i = 0; i < TEMP.length; i++) {
System.out.println(TEMP[i]);
}
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];
for (int i = 0; i < oldArray.length; i++) {
newArray[i]=oldArray[i];
}
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){
int length = 0;
int number1 = 1;
int number2 = 1;
int number3 = 1;
while (number3<max) {
if (number2 == 1) {
length = 2;
}else{
length++;
}
number3 = number1 + number2;
number1 = number2;
number2 = number3;
}
int Array[] = new int[length];
length = 1;
number1 = 1;
number2 = 1;
number3 = 1;
Array[0]= 1;
while (number3<max) {
number1 = number2;
number2 = number3;
Array[length]= number2;
length++;
number3 = number1 + number2;
}
return Array;
}
/**
* 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 length = 0;
int[] Primes;
if (max==3) {
Primes = new int[]{2};
}
boolean isPrimes;
for (int i = 2; i < max; i++) {
isPrimes = true;
for (int j = 2; j < i; j++) {
if (i%j==0 ) {
isPrimes = false;
break;
}
}
if (isPrimes) {
length++;
}
}
Primes = new int[length];
int num =0;
for (int i = 2; i < max; i++) {
isPrimes = true;
for (int j = 2; j < i; j++) {
if (i%j==0 ) {
isPrimes = false;
break;
}
}
if (isPrimes) {
Primes[num] = i;
num++;
}
}
return Primes;
}
/**
* 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[] PerfectNumbers;
int length = 0;
for (int i = 1; i < max; i++) {
int count=0;
for (int j = 1; j < i; j++) {
if (i%j==0) {
count = count + j;
}
}
if (i==count) {
length++;
}
}
PerfectNumbers = new int[length];
int num = 0;
for (int i = 1; i < max; i++) {
int count=0;
for (int j = 1; j < i; j++) {
if (i%j==0) {
count = count + j;
}
}
if (i==count) {
PerfectNumbers[num] = i;
num++;
}
}
return PerfectNumbers;
}
/**
* 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 strbur = new StringBuffer();
for (int i = 0; i < array.length; i++) {
if (i==array.length-1) {
strbur.append(array[i]);
}else {
strbur.append(array[i]).append(seperator);
}
}
return strbur.toString();
}
public static void main(String[] args) {
ArrayUtil arrryUtil = new ArrayUtil();
// int[] origin = {7, 9 , 30, 3};
// origin = arrryUtil.reverseArray(origin);
// for (int i = 0; i < origin.length; i++) {
// System.out.println(origin[i]);
// }
// int oldArr[]={1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5} ;
// int[] removeZero = arrryUtil.removeZero(oldArr);
// for (int i = 0; i < removeZero.length; i++) {
// System.out.println(removeZero[i]);
// }
// int[] a1 = {3, 5, 7,8};
// int[] a2 = {4, 5, 6,7};
// int[] mergeNew = arrryUtil.merge(a1, a2);
// for (int i = 0; i < mergeNew.length; i++) {
// System.out.println(mergeNew[i]);
// }
// int[] oldArray = {2,3,6};
// int[] grow = arrryUtil.grow(oldArray, 3);
// for (int i = 0; i < grow.length; i++) {
// System.out.println(grow[i]);
// }
// int[] fibonacci = arrryUtil.fibonacci(15);
// for (int i = 0; i < fibonacci.length; i++) {
// System.out.println(fibonacci[i]);
// }
// int[] Primes = arrryUtil.getPrimes(23);
// for (int i = 0; i < Primes.length; i++) {
// System.out.println(Primes[i]);
// }
// int[] PerfectNumbers = arrryUtil.getPerfectNumbers(2300);
// for (int i = 0; i < PerfectNumbers.length; i++) {
// System.out.println(PerfectNumbers[i]);
// }
int[] array= {3,8,9};
String seperator = "-";
String join = arrryUtil.join(array, seperator);
System.out.println(join);
}
}