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){
if(origin == null || origin.length <= 1){
return;
}
int tem = 0;
for(int i = 0, len = origin.length; i < len/2; i ++){
tem = origin[i];
origin[i] = origin[len - i - 1];
origin[len - i - 1] = tem;
}
System.out.println(join(origin, ","));
}
/**
* 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 withoutZeroSize = 0;
for(int i = 0, len = oldArray.length; i < len; i ++ ){
if( oldArray[i] != 0){
withoutZeroSize ++;
}
}
int[] newArray = new int[withoutZeroSize];
int point = 0;
for(int i = 0 ,len = oldArray.length; i < len; i ++ ){
if( oldArray[i] != 0){
newArray[point] = oldArray[i];
point ++;
}
}
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 point = 0;
int point2 = 0, point1 = 0;
int len1 = array1.length, len2 = array2.length;
int[] result = new int[len1 + len2];
while(point1 < len1 || point2 < len2){
if(point1 < len1 && point2 < len2){
if(array1[point1] <= array2[point2]){
result[point] = array1[point1];
point++;
point1++;
}else{
if(result[point - 1] == array2[point2]){
point2++;
}else{
result[point] = array2[point2];
point++;
point2++;
}
}
}else{
if(point1 < len1){
result[point] = array1[point1];
point++;
point1++;
}else{
result[point] = array2[point2];
point++;
point2++;
}
}
}
result = removeZero(result);
return result;
}
/**
* 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, len = oldArray.length; i < len; 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[] result = {1,1};
if(max <= 1){
return null;
}else if(max == 2){
return result;
}else{
result = grow(result, 10);
int index = 1;
while(true){
index ++;
if(result.length < index + 1){
result = grow(result, 10);
}
result[index] = result[index -1] + result[index - 2];
if(result[index] > max){
break;
}
}
}
return removeZero(result);
}
/**
* 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[] temArr = {2};
if(max < 2){
return null;
}else if (max == 2){
return temArr;
}else{
temArr = grow(temArr, 10);
int index = 0;
for(int i = 3;i < max; i ++){
boolean flag = true;
int iSqrt = (int) Math.sqrt(i);
for(int j = 0; j < index + 1; j ++){
if(iSqrt < temArr[j]){
break;
}
if(i % temArr[j] == 0){
flag = false;
break;
}
}
if(flag){
index ++;
if(temArr.length < index + 1){
temArr = grow(temArr, 30);
}
temArr[index] = i;
}
}
}
return removeZero(temArr);
}
/**
* 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[] result = new int[10];
int index = 0;
if(max < 6){
return null;
}
for (int n = 6; n <= max ; n ++){
int[] allFactors = getAllFactors(n);
int sum = 0;
for(int i = 0, len = allFactors.length; i < len; i ++){
sum += allFactors[i];
}
if(sum == n){
if(result.length < index + 1){
result = this.grow(result, 3);
}
result[index] = n;
index ++;
}
}
return removeZero(result);
}
public int[] getAllFactors(int n){
int[] result = new int[n];
int index = 0;
for(int i = 1; i < n; i++){
if(n % i == 0){
result[index] = i;
index ++;
}
}
return removeZero(result);
}
//分解因式算法
public int[] getPrimeFactors(int n){
int[] allPrimes = getPrimes(n);
int[] result = new int[allPrimes.length];
int index = 1;
result[0] = 1;
for(int i = 0, len = allPrimes.length; i < len; i ++){
int devide = n;
while(devide % allPrimes[i] == 0){
devide = devide / allPrimes[i];
result[index] = allPrimes[i];
index ++;
}
}
return this.removeZero(result);
}
/**
* 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 sb = new StringBuffer();
for(int i = 0, len = array.length; i < len; i ++){
if(i != 0){
sb.append(seperator);
}
sb.append(array[i]);
}
return sb.toString();
}
}