package assignment0226;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.TreeSet;
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) {
int mid = origin.length / 2;
for (int i = 0; i < mid; i++) {
int temp = origin[i];
int reversePosition = origin.length - 1;
origin[i] = origin[reversePosition];
origin[reversePosition] = 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
*/
public static int[] removeZero(int[] oldArray) {
int count = 0;
for (int i : oldArray) {
if (i != 0)
count++;
}
int[] newArray = new int[count];
int currentPos = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0)
newArray[currentPos++] = 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
*/
public static int[] merge(int[] array1, int[] array2) {
TreeSet<Integer> set = new TreeSet<>();
for (Integer integer : array1) {
set.add(integer);
}
for (Integer integer : array2) {
set.add(integer);
}
int[] result = new int[set.size()];
for (int i = 0; i < result.length; i++) {
result[i] = set.pollFirst();
}
return result;
}
/**
* 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) {
return Arrays.copyOf(oldArray, size);
}
/**
* 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];
List<Integer> fList = new ArrayList<>();
fList.add(1);
fList.add(1);
int last = fList.size() - 1;
while (fList.get(last) < max) {
fList.add(fList.get(last) + fList.get(last - 1));
last++;
}
int[] result = new int[fList.size() - 1];
for (int i = 0; i < result.length; i++) {
result[i] = fList.get(i);
}
return result;
}
/**
* 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) {
boolean[] isPrime = new boolean[max];
List<Integer> primes = new ArrayList<>();
for (int i = 0; i < isPrime.length; i++) {
isPrime[i] = true;
}
for (int i = 2; i * i < max; i++) {
for (int j = i; i * j < max; j++)
isPrime[i * j] = false;
}
for (int i = 2; i < isPrime.length; i++) {
if (isPrime[i])
primes.add(i);
}
int[] result = new int[primes.size()];
for (int i = 0; i < result.length; i++) {
result[i] = primes.get(i);
}
return result;
}
/**
* 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) {
int sum = 0;
ArrayList<Integer> perfectNumbers = new ArrayList<>();
for (int i = 1; i < max; i++) {
for (int j = 1; j < i; j++) {
if (i % j == 0) {
sum += j;
}
}
if (sum == i)
perfectNumbers.add(i);
sum = 0;
}
int[] result = new int[perfectNumbers.size()];
for (int i = 0; i < result.length; i++) {
result[i] = perfectNumbers.get(i);
}
return result;
}
/**
* 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 seperator
* @return
*/
public static String join(int[] array, String seperator) {
StringBuilder stringBuilder = new StringBuilder();
for (int i : array) {
stringBuilder.append(i + seperator);
}
stringBuilder.delete(stringBuilder.length() - seperator.length(), stringBuilder.length());
return stringBuilder.toString();
}
}