package com.coderising.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 size = origin.length;
int[] originReverse = new int[size];
for (int i = 0; i < size; i++) {
originReverse[i] = origin[size - 1 - i];
}
origin = originReverse;
}
/**
* 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 oldSize = oldArray.length;
int[] array = new int[oldSize];
int newSize = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
array[newSize] = oldArray[i];
newSize++;
}
}
if (oldSize == newSize) {
return array;
} else {
int[] temp = new int[newSize];
for (int i = 0; i < newSize; i++) {
temp[i] = array[i];
}
return temp;
}
}
/**
* 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 a[], int b[]) {
int lengthA = a.length;
int lengthB = b.length;
int c[] = new int[lengthA+lengthB];
int indexA = 0;
int indexB = 0;
int indexC = 0;
while (indexA < lengthA && indexB < lengthB) {
if (a[indexA] <= b[indexB]) {
if (indexC == 0 || c[indexC - 1] != a[indexA]) // 去重复
c[indexC++] = a[indexA];
indexA++;
} else {
if (indexC == 0 || c[indexC - 1] != b[indexB]) // 去重复
c[indexC++] = b[indexB];
indexB++;
}
}
while (indexA < lengthA) {
if (indexC == 0 || c[indexC - 1] != a[indexA]) // 去重复
c[indexC++] = a[indexA];
indexA++;
}
while (indexB < lengthB) {
if (indexC == 0 || c[indexC - 1] != b[indexB]) // 去重复
c[indexC++] = b[indexB];
indexB++;
}
int[] temp = new int[indexC];
for (int i = 0; i < indexC; i++) {
temp[i] = c[i];
}
return temp;
}
/**
* 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 oldArraySize = oldArray.length;
int[] temp = new int[oldArraySize + size];
for (int i = 0; i < oldArraySize; i++) {
temp[i] = oldArray[i];
}
return oldArray;
}
/**
* 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) {
int[] array = new int[20];
int i = 1;
int j = 1;
int sum = i+j;
array[0] = 1;
array[1] = 1;
array[2] = 2;
int k =3;
while(sum < max){
int temp = i;
i = j;
j = temp +j;
sum = i+j;
array[k++] = sum;
}
int[] temp = new int[k];
for (int p = 0; p < k; p++) {
temp[p] = array[p];
}
return temp;
}
/**
* 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[] a = new int[max];
int num = 0;
int j;
for (int i = 2; i < max; i++) {
for (j = 2; j <= Math.sqrt(i); j++) {
if (i % j == 0)
break;
}
if (j > Math.sqrt(i))
a[num++] = i;
}
int[] temp = new int[num];
for (int i = 0; i < temp.length; i++) {
temp[i] = a[i];
}
return temp;
}
/**
* 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[] a = new int[max];
int num = 0;
int j;
int sum = 0;
for (int i = 1; i < max; i++) {
sum = 0;
for (j = 1; j < i; j++) {
if (i % j == 0)
sum = sum + j;
}
if (sum == i) {
a[num++] = i;
}
}
int[] temp = new int[num];
for (int i = 0; i < temp.length; i++) {
temp[i] = a[i];
}
return temp;
}
/**
* 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 String join(int[] array, String seperator) {
int size = array.length;
StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < size - 1; i++) {
stringBuilder.append(array[i] + seperator);
}
stringBuilder.append(array[size - 1]);
return stringBuilder.toString();
}
}