package com.coding.basic.array;
import java.util.Arrays;
/**
* 数组工具类-第二次作业
* @author stackwei
* @date 2017/3/20
* @status ok
*/
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;
int[] temp;
length = origin.length;
temp = new int[length];
for (int i = 0; i < length; i++) {
temp[length - i - 1] = origin[i];
}
for (int i = 0; i < length; i++) {
origin[i] = temp[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 flag = 0;
int j = 0;
int length;
length = oldArray.length;
int[] newArray;
for (int i = 0; i < length; i++) {
if (oldArray[i] != 0) {
flag++;
}
}
newArray = new int[flag];
for (int i = 0; i < 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[] temp;
int[] array3;
int flag = 0;
int repeat = 0;
boolean boolea = true;
int length1 = array1.length;
int length2 = array2.length;
temp = new int[length1 + length2];
// 先把a1添加到temp
for (int i = 0; i < length1; i++) {
temp[i] = array1[i];
}
// 把a2中不重复的添加到temp
for (int i = 0; i < length2; i++) {
for (int j = 0; j < length1; j++) {
if (temp[j] == array2[i]) {
boolea = false;
repeat++;
}
}
if (boolea) {
temp[length1 + flag] = array2[i];
flag++;
}
boolea = true;
}
// 有重复就new一个数组长度减去重复的长度的a3,排序并返回
if (repeat != 0) {
array3 = new int[length1 + length2 - repeat];
for (int i = 0; i < (temp.length - repeat); i++) {
array3[i] = temp[i];
}
Arrays.sort(array3);
return array3;
}
// 无重复就排序并返回
Arrays.sort(temp);
return temp;
}
/**
* 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[] temp;
int length;
length = oldArray.length;
temp = new int[length + size];
for (int i = 0; i < length; i++) {
temp[i] = oldArray[i];
}
oldArray = null;
oldArray = temp;
return oldArray;
}
/**
* 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[] temp = new int[2];
int[] array;
int f1 = 1;
int f2 = 1;
int length = 2;
if (max <= 1) {
return null;
}
temp[0] = f1;
temp[1] = f2;
if (max == 2) {
return temp;
}
for (int i = 2; temp[i - 1] < max; i++) {
if ((f1 + f2) >= max)
break;
if (i + 1 > temp.length) {
temp = new ArrayUtil().grow(temp, 1);
}
temp[i] = f1 + f2;
f1 = temp[i - 1];
f2 = temp[i];
length++;
}
array = new int[length];
for (int i = 0; i < length; i++) {
array[i] = temp[i];
}
return array;
}
/**
* 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[] temp = new int[1];
boolean flag = true;
int i = 0;
if (max < 3)
return null;
for (int j = 2; j < max; j++) {
for (int k = 2; k <= Math.sqrt(j); k++) {
if (j % k == 0) {
flag = false;
break;
}
}
if (flag) {
if (i + 1 > temp.length)
temp = new ArrayUtil().grow(temp, 1);
temp[i] = j;
i++;
}
flag = true;
}
return temp;
}
/**
* 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[] temp = new int[1];
int i = 0;
if (max < 6)
return null;
for (int j = 1; j < max; j++) {
int total = 0;
for (int k = 1; k < j / 2 + 1; k++) {
if (j % k == 0)
total += k;
}
if (total == j) {
if (i + 1 > temp.length)
temp = new ArrayUtil().grow(temp, 1);
temp[i] = j;
i++;
}
}
return temp;
}
/**
* 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 || array.length ==0 || seperator == null)
return null;
StringBuilder sb = new StringBuilder();
int length = array.length;
for(int i=0;i<length;i++) {
sb.append(array[i]);
if(i != length-1)
sb.append(seperator);
}
return sb.toString();
}
}