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[] arr = new int[origin.length];
for (int i = 0; i < origin.length; i++) {
arr[i] = origin[origin.length-i-1];
}
for (int i = 0; i < arr.length; i++) {
origin[i] = arr[i];
}
}
/**
* 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 size = 0;
for (int i = 0; i < oldArray.length; i++) {
if(oldArray[i] != 0){
size ++;
}
}
int[] newArray = new int[size];
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[] totalArr = new int[array1.length + array2.length];
for (int i = 0; i < array1.length; i++) {
totalArr[i] = array1[i];
}
for (int j = array1.length,i = 0; i < array2.length; i++) {
totalArr[j] = array2[i];
j++;
}
//排序
for (int i = 0; i < totalArr.length; i++) {
for (int j = 0; j < totalArr.length-i-1; j++) {
if(totalArr[j] > totalArr[j+1]){
int temp = totalArr[j+1];
totalArr[j+1] = totalArr[j];
totalArr[j] = temp;
}
}
}
//去重
if(totalArr.length < 2){
return totalArr;
}
int size = 1;
for (int i = 0; i < totalArr.length-1; i++) {
if(totalArr[i] != totalArr[i+1]){
size ++ ;
}
}
int[] newArr = new int[size];
for (int i = 0,j = 0; i < totalArr.length-1; i++) {
if(totalArr[i] != totalArr[i+1]){
newArr[j] = totalArr[i];
j++;
}
}
newArr[size-1] = totalArr[totalArr.length-1];
return newArr;
}
/**
* 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 size = 0;
while(f(size) < max){
size++;
}
int[] arr = new int[size];
for (int i = 0; i < arr.length; i++) {
arr[i] = f(i);
}
return arr;
}
public int f(int n){
if(n == 0 || n ==1){
return 1;
}else{
return f(n-1)+f(n-2);
}
}
/**
* 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){
if(max <= 2){
return null;
}
int size = 1;
for (int i = 3; i < max; i++) {
int n = 2;
while(n < i){
if(i%n == 0){
break;
}
n++;
}
if(n == i){
size ++;
}
}
int[] arr = new int[size];
arr[0] = 2;
for (int i = 3,j = 1; i < max; i++) {
int n = 2;
while(n < i){
if(i%n == 0){
break;
}
n++;
}
if(n == i){
arr[j] = i;
j ++;
}
}
return arr;
}
/**
* 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){
if(max <= 5){
return null;
}
int size = 0;
for (int i = 6; i < max; i++) {
int sum = 0;
int n = 1;
while(n < i){
if(i % n == 0){
sum += n;
}
n++;
}
if(sum == i){
size ++;
}
}
int[] arr = new int[size];
for (int i = 6,j = 0; i < max; i++) {
int sum = 0;
int n = 1;
while(n < i){
if(i % n == 0){
sum += n;
}
n++;
}
if(sum == i){
arr[j] = i;
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 String join(int[] array, String seperator){
String str = "";
for (int i = 0; i < array.length; i++) {
if(i != array.length-1){
str += (array[i] + seperator);
}else{
str += array[i];
}
}
return str;
}
}