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 length = origin.length;
for (int i = 0, j = length / 2; i < j; i++) {
int temp = origin[i];
origin[i] = origin[length - i - 1];
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 newLength = 0;
for (int t : oldArray){
newLength += t != 0 ? 1 : 0;
}
int[] newArr = new int[newLength];
int i = 0;
for (int t : oldArray) {
if (t != 0) {
newArr[i] = t;
i++;
}
}
return newArr;
}
/**
* 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 array1Length = array1.length;
int array2Length = array2.length;
int[] newArray=new int[array1Length+array2Length];
int i = 0, j = 0, index = 0;
while (i < array1.length && j < array2.length) {
if (array1[i] <= array2[j]) {
if (array1[i] == array2[j]) {
j++;
} else {
newArray[index] = array1[i];
i++;
index++;
}
} else {
newArray[index] = array2[j];
j++;
index++;
}
}
//当array2为null
while (i < array1.length) {
newArray[index] = array1[i];
index++;
i++;
}
//当array1为null
while (j < array2.length) {
newArray[index] = array2[j];
j++;
index++;
}
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 oldArrLength = oldArray.length;
int newArrLength = oldArrLength + size;
int[] newArray = new int[newArrLength];
System.arraycopy(oldArray,0,newArray,0,newArrLength);
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[] fibonacci = new int[max];
if (max <=1) {
return null;
} else if (max == 2) {
fibonacci[0] = 1;
fibonacci[1] = 1;
} else {
for (int i =2;i<max;i++) {
fibonacci[0] = 1;
fibonacci[1] = 1;
fibonacci[i] = fibonacci[i - 1] + fibonacci[i - 2];
}
}
return fibonacci;
}
/**
* 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){
boolean isPrimes = false;
int[] primes=new int[max];
int index = 0;
for (int i=2;i<max;i++) {
//判断素数
for (int j=2;j<Math.sqrt(i);j++) {
if (i % j == 0) {
isPrimes = false;
} else isPrimes = true;
}
if (isPrimes) {
primes[index] = i;
index++;
}
}
return primes;
}
/**
* 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[] perfectNum = new int[max];
int index = 0;
for (int i=0;i<max;i++) {
int sum = 0;
for (int j=0;j<max;j++) {
if (i % j == 0) {
sum = sum + i;
}
if (i == j) {
perfectNum[index] = sum;
index++;
}
}
}
return perfectNum;
}
/**
* 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){
String[] newArray = new String[2*array.length-1];
int i = 0, j = i + 1;
for (int k=0;k<array.length;k++) {
newArray[2*k] = String.valueOf(array[k]);
newArray[2*k+1] = seperator;
}
return newArray.toString();
}
}