package com.coderising;
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) {
for (int i = 0; i < origin.length / 2; i++) {
int 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
* @return
*/
public static 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];
int j = 0;
for (int i = 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
*/
public static int[] merge(int[] array1, int[] array2) {
int[] newArr = new int[array1.length + array2.length];
int index1 = 0;
int index2 = 0;
int z = 0;
for (int i = 0; i < array2.length; i++) {
if (array1[index1] - array2[index2] > 0) {
newArr[z++] = array2[index2];
}
}
return null;
}
public static int[] merge1(int[] array1, int[] array2) {
int size = 0;
for (int i = 0; i < array1.length; i++) {
for (int j = 0; j < array2.length; j++) {
if (array1[i] == array2[j]) {
size++;
}
}
}
int[] newArr = new int[array1.length + array2.length - size];
int z = 0;
for (int i = 0; i < array1.length; i++) {
for (int j = 0; j < array2.length; j++) {
if (array1[i] < array2[j]) {
newArr[z++] = array1[i];
break;
} else if (array1[i] > array2[j]) {
newArr[z++] = array2[j];
break;
} else if (array1[i] == array2[j]) {
newArr[z++] = array1[i];
break;
}
}
}
System.out.println(size);
return null;
}
/**
* 把一个已存满数据的数组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
*
* @param args
*/
public static int[] grow(int[] oldArray, int size) {
int[] newArray = new int[oldArray.length + size];
System.arraycopy(oldArray, 0, newArray, 0, size);
return newArray;
}
/**
* 裴波那契数列为:1,1,2,3,5,8,13,21。。。 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 args
*/
public static int[] fibonacci(int max) {
if (max == 1) {
int[] arr = new int[0];
return arr;
}
int x = 1;
int y = 1;
int temp = 0;
int size = 0;
while (true) {
temp = y;
y = x + y;
x = temp;
if (y >= max) {
break;
}
size++;
}
int[] array = new int[size + 2];
array[0] = 1;
array[1] = 1;
int j = 2;
int m = 1;
int n = 1;
int temp1 = 1;
while (true) {
temp1 = n;
n = m + n;
m = temp1;
if (n >= max) {
break;
}
size++;
array[j++] = n;
}
return array;
}
/**
* 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
*/
public static int[] getPrimes(int max) {
int i = 0;
for (int j = 2; j < max; j++) {
boolean b = false;
for (int j2 = 2; j2 < j; j2++) {
if (j % j2 == 0) {
b = true;
break;
}
}
if (b == false) {
i++;
}
}
int[] arr = new int[i];
int z = 0;
for (int j = 2; j < max; j++) {
boolean b = false;
for (int j2 = 2; j2 < j; j2++) {
if (j % j2 == 0) {
b = true;
break;
}
}
if (b == false) {
arr[z++] = j;
}
}
return arr;
}
/**
* 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) {
for (int i = 0; i < array.length; i++) {
if (i == 0) {
seperator = array[i] + "";
} else {
seperator = seperator + "-" + "" + array[i] + "";
}
}
return seperator;
}
/**
* 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){
return null;
}
public static void main(String[] args) {
int[] a = { 1, 3, 5, 6 };
int[] b = { 2, 3, 4, 5, 7 };
int[] ab = ArrayUtil.fibonacci(30);
int array[] = { 1, 2, 3, 4 };
String s = ArrayUtil.join(array, new String());
System.out.println(s);
}
}