package com.coderising.week02.array;
import java.util.Stack;
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){
Stack stack = new Stack();
for(int i = 0; i < origin.length; i++ ) {
stack.push(origin[i]);
}
for(int i = 0; i < origin.length; i++ ) {
int j = (int) stack.pop();
origin[i] = j;
}
}
/**
* 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){
if (oldArray == null) {
throw new IllegalArgumentException();
}
//算出oldArr数组中0的个数
int zeronum = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] == 0) {
zeronum++;
}
}
int j = 0; //新数组的索引
int[] newArray = new int[oldArray.length - zeronum];
//遍历旧数组
for (int i = 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 array1length = array1.length;
int array2length = array2.length;
int repeat = 0; //初始化两个数组中重复数字的个数为0
for (int i = 0; i < array1length; i++) {
for (int j = 0; j < array2length; j++) {
if (array1[i] == array2[j]) {
repeat++;
}
}
}
//合并后的数组长度为两个数组长度相加减去重复数字的个数
int[] mergearray = new int[array1length+array2length-repeat];
int index1 = 0; //array1数组的当前索引
int index2 = 0; //array2数组的当前索引
int indexMerge = 0; //mergearray数组的当前索引
//循环,只要array1和array2都没有循环完就一直循环
while (index1 < array1length && index2 < array2length) {
//如果当前array1[index1]比array2[index2]小,则将mergearray[indexMerge]元素置为array1[index1]
//同时index1++,indexMerge++
if (array1[index1] < array2[index2]) {
mergearray[indexMerge++] = array1[index1++];
}
//如果当前array1[index1]比array2[index2]大,则将mergearray[indexMerge]元素置为array2[index2]
//同时index2++,indexMerge++
if (array1[index1] > array2[index2]) {
mergearray[indexMerge++] = array2[index2++];
}
//如果当前array1[index1]等于array2[index2],则将mergearray[indexMerge]元素置为array1[index1]
//同时index1++,index2++,indexMerge++
else {
mergearray[indexMerge] = array1[index1];
index1++;
index2++;
indexMerge++;
}
}
//上个循环能够结束,说明array1已经循环完或array2已经循环完
//下述两个循环只能有一个满足循环条件
//只要array1没有循环完,就把array1中剩下的元素依次放入mergearray中
while (index1 < array1length) {
mergearray[indexMerge++] = array1[index1++];
}
//只要array2没有循环完,就把array2中剩下的元素依次放入mergearray中
while (index2 < array2length) {
mergearray[indexMerge++] = array2[index2++];
}
return mergearray;
}
/**
* 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 oldLength = oldArray.length;
int[] newArray = new int[oldLength+size];
System.arraycopy(oldArray, 0, newArray, 0, oldLength);
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 static int[] fibonacci(int max){
int count = 0; //数列项数
//循环得到数组的长度
while (fiboindex(count) < max) {
count++;
}
int[] fiboArray = new int[count];
if(max <= 1) {
fiboArray = new int[] {};
} else {
for (int i = 0; i < count; i++) {
fiboArray[i] = fiboindex(i);
}
}
return fiboArray;
}
//通过递归的方式推导斐波那契数列
public static int fiboindex(int n) {
if (n < 2) {
return 1;
}
else {
return fiboindex(n-1) + fiboindex(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) {
throw new IllegalArgumentException();
}
int i,j;
int count = 0;
int[] maxArray = new int[max];
for(i = 2; i < max; i++) {
for (j = 2; j <= i; j++) {
if (i % j == 0) {
break;
}
}
if(j >= i) {
maxArray[count]=i;
count++;
}
}
int[] primeArray = new int[count];
System.arraycopy(maxArray, 0, primeArray, 0, count);
return primeArray;
}
/**
* 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 < 0) {
throw new IllegalArgumentException();
}
int i,j;
int count = 0;
int[] maxArray = new int[max];
for(i = 1; i < max; i++) {
int sum = 0;
for(j = 1; j < i; j++) {
if (i % j == 0) {
sum += j;
}
}
if (sum == i) {
maxArray[count] = i;
count++;
}
}
int[] perfectArray = new int[count];
System.arraycopy(maxArray, 0, perfectArray, 0, count);
return perfectArray;
}
/**
* 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){
if (array == null) {
throw new IllegalArgumentException();
}
if (array.length == 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < array.length; i++) {
sb.append(array[i]);
if(i != array.length - 1) {
sb.append(seperator);
}
}
return sb.toString();
}
}