package com.github.FelixCJF.coding2017.coderising.array;
import java.util.ArrayList;
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 array[] = new int[origin.length];
for (int i = 0; i<origin.length; i++) {
array[i] = origin[origin.length - 1 - i];
}
System.arraycopy(array, 0, origin, 0, origin.length);
}
/**
* 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 zero = 0;
//统计原来数组0的个数
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] == 0) {
zero ++;
}
}
int[] newArr = new int[oldArray.length - zero];
int count = 0;
for (int i = 0; i<oldArray.length; i ++) {
if (oldArray[i] != 0) {
newArr[count] = oldArray[i];
count ++;
}
}
return newArr;
}
/**
* 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){
//创建新数组
int length1 = array1.length;
int length2 = array2.length;
int length = length1 + length2;
int[] newArr = new int[length];
//插入
for (int i = 0; i < length1; i++) {
newArr[i] = array1[i];
}
for (int i = 0; i < length2; i++) {
newArr[length1 - 1 + i] = array2[i];
}
//排序
for (int i = 0; i < length-1; i ++){
if (newArr[i] > newArr[i+1]) {
int temp = newArr[i];
newArr[i] = newArr[i+1];
newArr[i + 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 newArr[] = new int[oldArray.length + size];
System.arraycopy(oldArray, 0, newArr, 0, oldArray.length);
return newArr;
}
/**
* 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[] newArr;
int f1 = 0;
int f2 = 1;
int f = 0;
if (max < 2) {
return newArr = new int[0];
}
ArrayList list = new ArrayList();
for (int i = 2; f < max; i++) {
list.add(f2);
f = f1 + f2;
f1 = f2;
f2 = f;
}
newArr = new int[list.size()];
for (int i = 0; i < newArr.length; i++) {
newArr[i] = (int) list.get(i);
}
return newArr;
}
/**
* 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){
ArrayList list = new ArrayList();
for (int i = 1; i < max; i++) {
if (isPrime(i)) {
list.add(i);
}
}
int[] newArr = new int[list.size()];
for (int i = 0; i < newArr.length; i++) {
newArr[i] = (int) list.get(i);
}
return newArr;
}
//判断是否为素数
private boolean isPrime(int a) {
boolean flag = true;
if (a < 2) {// 素数不小于2
return false;
} else {
for (int i = 2; i <= Math.sqrt(a); i++) {
if (a % i == 0) {// 若能被整除,则说明不是素数,返回false
flag = false;
break;// 跳出循环
}
}
}
return flag;
}
/**
* 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[] newArr;
if (max == 0) {
return newArr = new int[0];
}
ArrayList list = new ArrayList();
for (int i = 1; i < max; i++) {
if (isWanshu(i)) {
list.add(i);
}
}
newArr = new int[list.size()];
for (int i = 0; i < newArr.length; i++) {
newArr[i] = (int) list.get(i);
}
return newArr;
}
//判断一个数是不是完数
private boolean isWanshu(int n)
{
boolean flag=false;
int i,sum=0;
for(i=1;i<=n/2;i++)
{
if(n%i==0)
{
sum+=i;
}
}
if(sum==n)
{
flag=true;
}
return flag;
}
/**
* 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 string = "";
for (int i = 0; i < array.length; i++) {
if (i == array.length - 1) {
string += array[i];
} else {
string += array[i] + seperator;
}
}
return string;
}
}