package array;
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) {
int temp;
for (int 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
* @return
*/
public int[] removeZero(int[] oldArray) {
int count = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] == 0) {
count++;
}
}
int[] newArray = new int[oldArray.length - count];
int flag = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArray[flag] = oldArray[i];
flag++;
} else {
continue;
}
}
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 int[] merge(int[] array1, int[] array2) {
/*
int[] temp = new int[array1.length + array2.length];
int count = array1.length;
int point = array1.length;
for (int i = 0; i < array1.length; i++) {
temp[i] = array1[i];
}
boolean flag = true;
for (int i = 0; i < array2.length; i++) {
for (int j = 0; j < array1.length; j++) {
if (array1[j] == array2[i]) {
flag = false;
}
}
if (flag) {
temp[count]=array2[i];
count++;
}
flag = true;
}
*/
if (array1.length == 0) {
return array2;
}
if (array2.length == 0) {
return array1;
}
int[] temp = new int[array1.length + array2.length];
int ap = 0;
int bp = 0;
int count = 0;
while (ap < array1.length && bp < array2.length) {
if (array1[ap] == array2[bp]) {
temp[count] = array1[ap];
ap++;
bp++;
count++;
} else if (array1[ap] > array2[bp]) {
temp[count] = array2[bp];
bp++;
count++;
} else {
temp[count] = array1[ap];
ap++;
count++;
}
}
if (ap == array1.length) {
for (int i = bp; i < array2.length; i++) {
temp[count] = array2[i];
count++;
}
} else if (bp == array2.length) {
for (int i = ap; i < array1.length; i++) {
temp[count] = array1[i];
count++;
}
}
int array3[] = new int[count];
System.arraycopy(temp, 0, array3, 0, count);
return array3;
/*int[] temp = new int[array2.length];
boolean flag = true;
int count = 0;
for (int i = 0; i < array2.length; i++) {
for (int j = 0; j < array1.length; j++) {
if (array2[j] == array1[i]) {
flag = false;
}
}
if (flag) {
temp[count] = array2[i];
count++;
}
}
if (count == 0) {
return array1;
}
int ap = 0; //数组1的指针
int bp = 0; //数组2的指针
int[] array3 = new int[count + array1.length];
for (int i = 0; i < array3.length; i++) {
if (array1[ap] > array2[bp]) {
array3[i] = array2[bp];
bp++;
}else {
array3[i] = array1[ap];
ap++;
}
}
*/
}
/**
* 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) {
int[] newArray = new int[oldArray.length + size];
for (int i = 0; i < oldArray.length; i++) {
newArray[i] = oldArray[i];
}
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 null;
}
int a = 1;
int b = 1;
int[] result = {1, 1};
int[] temp;
while (b < max) {
b = a + b;
a = b - a;
temp = result;
result = new int[result.length + 1];
for (int i = 0; i < temp.length; i++) {
result[i] = temp[i];
}
result[result.length - 1] = b;
}
temp = result;
result = new int[result.length - 1];
for (int i = 0; i < result.length; i++) {
result[i] = temp[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 int[] getPrimes(int max) {
/* int[] result = null;
int[] temp = null;
if (max < 2) {
return null;
}
boolean flag = true;
for (int i = 2; i < max; i++) {
for (int j = 2; j * j < i; j++) {
if (i % j == 0) {
flag = false;
}
}
if (flag) {
if (result == null) {
result = new int[1];
result[0] = i;
} else {
temp = result;
result = new int[result.length + 1];
for (int j = 0; j < temp.length; j++) {
result[j] = temp[j];
}
result[result.length - 1] = i;
}
}
flag = true;
}
return result;*/
if (max < 2) {
return null;
}
int[] result = {2};
int[] temp ;
boolean flag = true;
for (int i = 3; i < max; i++) {
for (int j = 2; j * j <= i; j++) {
if (i % j == 0){
flag = false;
}
}
if (flag) {
temp=result;
result=new int[temp.length+1];
System.arraycopy(temp,0,result,0,temp.length);
result[result.length-1]=i;
}
flag=true;
}
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 int[] getPerfectNumbers(int max) {
int[] result = {};
int[] temp = null;
int count = 0;
for (int i = 1; i < max; i++) {
for (int j = 1; j < i; j++) {
if (i % j == 0) {
count += j;
}
}
if (count == i) {
temp = result;
result = new int[temp.length + 1];
for (int j = 0; j < temp.length; j++) {
result[j] = temp[j];
}
result[result.length - 1] = 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 String join(int[] array, String seperator) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < array.length - 1; i++) {
sb.append(array[i]);
sb.append(seperator);
}
sb.append(array[array.length - 1]);
return sb.toString();
}
}