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 static void reverseArray(int[] origin){
for(int i = 0;i < origin.length/2; i++){
int x = origin[i];
origin[i] = origin[origin.length - i -1];
origin[origin.length - i -1] = x;
}
}
/**
* 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 static int[] removeZero(int[] oldArray){
int zeroCount = 0;
for(int i = 0;i < oldArray.length; i++){
if(oldArray[i] == 0){
zeroCount++;
}
}
int[] newArr = new int[oldArray.length - zeroCount];
int index = 0;
for(int i = 0;i < oldArray.length; i++){
if(oldArray[i] != 0){
newArr[index] = oldArray[i];
index++;
}
}
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 static int[] merge(int[] array1, int[] array2){
//先对数组进行去重,记录重复的索引,后将两个数组合并,再进行排序
int[] repeatedNum = new int[array1.length + array2.length];
int repeatedCount = 0;
for(int i = 0;i < array1.length; i++){
for(int j = 0;j < array2.length; j++){
if(array1[i] == array2[j]){
repeatedNum[repeatedCount] = array1[i];
repeatedCount++;
}
}
}
int [] combineArr = new int[array1.length + array2.length - repeatedCount];
for(int i = 0;i < array1.length; i++){
combineArr[i] = array1[i];
}
for(int i = 0;i < array2.length; i++){
int index = array1.length -1;
boolean same = false;
for(int j = 0;j < repeatedNum.length; j++){
if(array2[i] == repeatedNum[j]){
same = true;
}
}
if(!same){
index += 1;
combineArr[index] = array2[i];
}
}
//冒泡排序
for(int i = 0;i < combineArr.length;i++){
for(int j = i + 1;j < combineArr.length;j++){
if(combineArr[i] > combineArr[j]){
int x = combineArr[i];
combineArr[i] = combineArr[j];
combineArr[j] = x;
}
}
}
return combineArr;
}
/**
* 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 static int[] grow(int [] oldArray, int size){
int[] newArr = new int[oldArray.length + size];
for(int i = 0;i < oldArray.length; i++){
newArr[i] = oldArray[i];
}
return newArr;
}
/**
* 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){
if(max == 1){
return null;
}else{
int length = 0;
int dataBefore = 0;
int dataAfter = 1;
while(dataAfter < max){
int date = dataAfter;
dataAfter = dataAfter + dataBefore;
dataBefore = date;
length++;
}
int index = 0;
int[] result = new int[length];
dataBefore = 0;
dataAfter = 1;
while(dataAfter < max){
result[index] = dataAfter;
int date = dataAfter;
dataAfter = dataAfter + dataBefore;
dataBefore = date;
index ++;
}
return 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 static int[] getPrimes(int max){
int i = 1;
int length = 0;
while(i < max){
i++;
int search = 1;
}
return null;
}
/**
* 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 static int[] getPerfectNumbers(int max){
return null;
}
/**
* 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 static String join(int[] array, String seperator){
StringBuilder sb = new StringBuilder();
for(int i=0 ;i < array.length; i++){
sb.append(String.valueOf(array[i]));
if(i != array.length - 1){
sb.append(seperator);
}
}
return sb.toString();
}
public static void main(String[] args) {
/*int[] a = {7, 9 , 30, 3};
reverseArray(a);
for (int i : a) {
System.out.print(i+",");
}*/
/*int[] oldArr = {1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5} ;
int[] newArr = removeZero(oldArr);
for (int i : newArr) {
System.out.print(i+",");
}*/
/*int[] a1 = {3, 5, 7,8};
int[] a2 = {4, 5, 6,7};
int[] merge = merge(a1,a2);
for (int i : merge) {
System.out.print(i+",");
}*/
/*int[] oldArray = {2,3,6};
int size = 3;
int[] newArr = grow(oldArray, size);
for (int i : newArr) {
System.out.print(i+",");
}*/
/*int[] array= {3,8,9};
String seperator = "-";
String join = join(array, seperator);
System.out.println(join);*/
int[] fibonacci = fibonacci(15);
for (int i : fibonacci) {
System.out.print(i+",");
}
}
}