package main;
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 static void reverseArray(int[] origin) {
if (null == origin) {
return;
}
for (int temp = 0, i = 0; i < origin.length / 2; i++) {
temp = origin[i];
origin[i] = origin[origin.length - i - 1];
origin[origin.length - i - 1] = 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
* @param newArray
* @return
*/
public static int[] removeZero(int[] oldArray) {
if (null == oldArray) {
return null;
}
int counter = 0;
int[] newArray = new int[oldArray.length];
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArray[counter++] = oldArray[i];
}
}
return Arrays.copyOf(newArray, counter);
}
/**
* 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 static int[] merge(int[] array1, int[] array2) {
if (null == array1) {
return array2;
}
if (null == array2) {
return array1;
}
if (0 == array1.length && 0 == array2.length) {
return new int[0];
}
Arrays.sort(array1);
Arrays.sort(array2);
int[] array = new int[array1.length + array2.length];
System.arraycopy(array1, 0, array, 0, array1.length);
System.arraycopy(array2, 0, array, array1.length, array2.length);
int counter = 0;
for (int i = 0; i < array.length - 1 - counter; i++) {
for (int j = i + 1; j < array.length - counter; j++) {
if (array[i] == array[j]) {
int k = 0;
for (k = i; k < array.length - 1; k++) {
array[k] = array[k + 1];
}
i--;
counter++;
break;
}
}
}
array = Arrays.copyOf(array, array.length - counter);
Arrays.sort(array);
return array;
}
/**
* 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 static int[] grow(int[] oldArray, int size) {
if (null == oldArray) {
return null;
}
int[] array = new int[oldArray.length + size];
for (int i = 0; i < oldArray.length; i++) {
array[i] = oldArray[i];
}
return array;
}
/**
* 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 static int[] fibonacci(int max) {
if (max <= 1) {
return new int[0];
}
int[] array = new int[max];
array[0] = 1;
array[1] = 1;
int i = 1;
do {
i++;
array[i] = array[i - 1] + array[i - 2];
} while (array[i] < max);
return Arrays.copyOf(array, i);
}
/**
* 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 static int[] getPrimes(int max) {
if (max < 3) {
return new int[0];
}
int[] array = new int[max];
int counter = 0;
for (int i = 2; i < max; i++) {
int j = 2;
for (; j * j <= i; j++) {
if (i % j == 0) {
break;
}
}
if (j * j > i) {
array[counter] = i;
counter++;
}
}
return Arrays.copyOf(array, counter);
}
/**
* 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 static int[] getPerfectNumbers(int max) {
if (max < 7) {
return new int[0];
}
int[] array = new int[2];
int counter = 0;
for (int i = 6; i < max; i++) {
int sum = 0;
for (int j = 1; j <= i / 2; j++) {
if (i % j == 0) {
sum = sum + j;
}
}
if (sum == i) {
if (counter == array.length) {
array = ArrayUtil.grow(array, 2);
}
array[counter++] = i;
}
}
return Arrays.copyOf(array, counter);
}
/**
* 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 static String join(int[] array, String seperator) {
if (null == array) {
return null;
}
if (array.length == 0) {
return "";
}
StringBuffer sb = new StringBuffer();
for (int i = 0; i < array.length; i++) {
sb.append(seperator).append(array[i]);
}
return sb.toString().substring(seperator.length());
}
}