package com.coderising.array;
import java.util.Iterator;
import java.util.TreeSet;
import org.junit.Test;
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 leon1900
*/
public void reverseArray(int[] origin){
for (int i = 0; i < origin.length/2; i++) {
int temp = origin[origin.length-1-i];
origin[origin.length-1-i] = 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 leon1900
*/
public int[] removeZero(int[] oldArray){
int size = 0;
for (int i = 0; i < oldArray.length; i++) {
if(oldArray[i] !=0 ){
size++;
}
}
int [] newArray = new int [size];
for (int i = 0,j = 0; i < oldArray.length; i++) {
if(oldArray[i]!=0){
newArray[j++] = oldArray[i];
}
}
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 leon1900
*/
public int[] merge(int[] array1, int[] array2){
TreeSet<Integer> ts = new TreeSet<>();
for (int i = 0; i < array1.length; i++) {
ts.add(array1[i]);
}
for (int i = 0; i < array2.length; i++) {
ts.add(array2[i]);
}
int size = ts.size();
int[] newArr = new int[size];
int i = 0;
for (Integer integer : ts) {
newArr[i++] = integer;
}
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 leon1900
*/
public int[] grow(int [] oldArray, int size){
int newsize = oldArray.length+size;
int [] newArr = new int[newsize];
for (int i = 0; i < oldArray.length; i++) {
newArr[i] = oldArray[i];
}
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 leon1900
*/
public int[] fibonacci(int max){
int size = 1;
while(fun(size)<max){
size++;
}
int [] arr = new int [size-1];
for (int j=0,i = 1; i < max; i++) {
if(fun(i)<max){
arr[j++] = fun(i);
}else {
return arr;
}
}
return null;
}
/**
* 求给定数的斐波拉契数
* @param max
* @return leon1900
*/
private int fun(int n) {
if(n==1||n==2){
return 1;
}else {
return fun(n-1)+fun(n-2);
}
}
/**
* 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 leon1900
*/
public int[] getPrimes(int max){
int size = 0;
for (int i = 1; i < max; i++) {
if(primeNumber(i)){
size++;
}
}
int []arr = new int[size];
for (int j=0,i = 1; i < max; i++) {
if(primeNumber(i)){
arr[j++]=i;
}
}
return arr;
}
/**
* 判断是否为素数
* @param num
* @return leon1900
*/
public boolean primeNumber(int num){
boolean flag = true;
if(num==1) flag = false;
for (int j = 2; j <= num/2; j++) {
if(num % j == 0){
flag = false;
break;
}
}
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 size=0;
for (int i = 1; i < max; i++) {
if(perfectNumber(i)){
size++;
}
}
int[] arr = new int[size];
for (int j=0,i = 1; i < max; i++) {
if(perfectNumber(i)){
arr[j++] = i;
}
}
return arr;
}
public boolean perfectNumber(int num){
boolean flag = false;
int sum = 0;
//完数 除了1 跟本身的因子
for (int i = 1; i <= num/2; i++) {
if(num%i==0){
sum+=i;
}
}
if(sum==num)
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 leon1900
*/
public String join(int[] array, String seperator){
StringBuilder sBuilder =new StringBuilder();
for (int i = 0; i < array.length; i++) {
if(i==array.length-1){
sBuilder.append(array[i]);
}else{
sBuilder.append(array[i]+seperator);
}
}
return sBuilder.toString();
}
}