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 void reverseArray(int[] origin){
if(origin == null || origin.length == 0) {
return;
}
for(int i=0, j = origin.length-1; i < j; i++,j++) {
int t = origin[i];
origin[i] = origin[j];
origin[j] = t;
}
}
/**
* 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){
if(oldArray == null) {
return null;
}
int counter = 0; // to count how many non-zero numbers
int b[] = new int[oldArray.length];
//count non-zero numbers
for(int i=0; i < oldArray.length; i++) {
if(oldArray[i] != 0)
{
b[counter++] = oldArray[i];
}
}
int newArray[] = new int[counter];
return Arrays.copyOf(b,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 int[] merge(int[] array1, int[] array2){
if(array1 == null && array2 == null) {
return null;
}
int [] newArray = new int[array1.length + array2.length];
int i = 0;
int j = 0;
int counter = 0;
while(i<array1.length && j<array2.length)
{
if(array1[i] < array2[j])
{
newArray[counter++] = array1[i++];
}
else if(array1[i] > array2[j])
{
newArray[counter++] = array2[j++];
}
else if(array1[i] == array2[j])
{
newArray[counter++] = array2[j];
i++;
j++;
}
}
while(i==array1.length && j<array2.length)
{
newArray[counter++] = array2[j++];
}
while(j==array2.length && i<array1.length)
{
newArray[counter++] = array1[i++];
}
int[] newArray1 = new int[counter];
return Arrays.copyOf(newArray1,counter);
}
/**
* 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){
if(oldArray == null)
return null;
if(size < 0)
throw new IndexOutOfBoundsException("size is smaller than 0");
int[] newArray = new int[oldArray.length + size];
System.arraycopy(oldArray, 0, newArray, 0, oldArray.length);
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){
if(max == 1) {
return new int[0];
}
if(max == 2) {
return new int[] {1,1};
}
int[] a = new int[max];
a[0] = 1;
a[1] = 1;
int counter = 2;
for(int i = 2; i < max; i++) {
a[i] = a[i - 1] + a[i - 2];
if(a[i] >= max) {
break;
} else {
counter++;
}
}
return Arrays.copyOf(a, counter);
}
/**
* 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){
if(max < 3)
return new int[0];
int[] array = new int[max];
int counter = 0;
for(int n = 2; n < max; n++) {
if (isPrime(n)) {
array[counter++] = n;
}
}
return Arrays.copyOf(array, counter);
}
private boolean isPrime(int n) {
int i = 2;
while (i < n) {
if (n % i == 0) {
break;
}
if (n % i != 0) {
i++;
}
}
return i == n;
}
/**
* 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){
if(max <= 0) return new int[0];
int[] array = new int[max];
int counter = 0;
for(int n = 2; n < max; n++)
{
int sum = 0;
for(int i = 1; i < n; i++)
{
if(n % i == 0)
{
sum += i;
}
}
if(sum == n)
{
array[counter++] = n;
}
}
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
* @return
*/
public String join(int[] array, String seperator) {
if (array == null) return null;
if(array.length == 0) return "";
StringBuilder buffer = new StringBuilder();
for (int i = 0; i < array.length; i++) {
buffer.append(array[i]);
if (i < array.length - 1) {
buffer.append(seperator);
}
}
return buffer.toString();
}
}