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 l = origin.length, n;
for (int i = 0; i < l / 2; i++) {
n = origin[i];
origin[i] = origin[l - 1 - i];
origin[l - 1 - i] = n;
}
}
/**
* 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 n = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] == 0)
n++;
}
int[] newArray = new int[oldArray.length - n];
for (int i = 0, j = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArray[j] = oldArray[i];
j++;
}
}
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 n = 0; // 重复的整数个数
for (int i = 0, j = 0; i < array1.length && j < array2.length;) {
int z = array1[i] - array2[j];
if (z < 0)
i++;
else if (z > 0)
j++;
else {
i++;
j++;
n++;
}
}
int[] newArray = new int[array1.length + array2.length - n];
for (int i = 0, j = 0, k = 0; i < array1.length && j < array2.length;) {
int z = array1[i] - array2[j];
if (z < 0) {
newArray[k] = array1[i];
k++;
i++;
} else if (z > 0) {
newArray[k] = array2[j];
k++;
j++;
} else {
newArray[k] = array1[i];
k++;
i++;
j++;
}
// 判断循环是否即将结束,若结束则将另外一个数组未比较的元素放入新数组
if (i >= array1.length) {
for (int a = j; a < array2.length; a++, k++) {
newArray[k] = array2[a];
}
}
if (j >= array2.length) {
for (int a = i; a < array1.length; a++, k++) {
newArray[k] = array1[a];
}
}
}
return newArray;
}
/**
* 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) {
int n = 0;
for (int i = 1, j = 1, k; i < max; n++) {
k = i + j;
i = j;
j = k;
}
int[] newArray = new int[n];
if (n > 1) {
newArray[0] = 1;
newArray[1] = 1;
}
for (int i = 2; i < n; i++) {
newArray[i] = newArray[i - 1] + newArray[i - 2];
}
return newArray;
}
/**
* 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 n = 0;
for (int i = 2; i < max; i++) {
if (primeNumber(i))
n++;
}
int[] newArray = new int[n];
for (int i = 2, j = 0; i < max; i++) {
if (primeNumber(i)) {
newArray[j++] = i;
}
}
return newArray;
}
// 判断是否为素数
public boolean primeNumber(int number) {
for (int i = 2; i < number; i++) {
if (number % i == 0)
return false;
}
return true;
}
/**
* 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 n = 0;
for (int i = 6; i < max; i++) {
if (perfectNumber(i))
n++;
}
int[] newArray = new int[n];
for (int i = 6, j = 0; i < max; i++) {
if (perfectNumber(i)) {
newArray[j++] = i;
}
}
return newArray;
}
// 判断是否为完数
public boolean perfectNumber(int number) {
int n = 0;
for (int i = 1; i <= number / 2; i++) {
if (number % i == 0)
n += i;
}
if (number != n)
return false;
return true;
}
/**
* 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) {
StringBuffer s = new StringBuffer();
for (int i = 0; i < array.length; i++) {
s.append(String.valueOf(array[i]));
s.append("-");
}
String str = s.substring(0, s.length() - 1);
return str;
}
}