package io.github.vxzh.struts;
import java.util.ArrayList;
import java.util.List;
/**
* Created by vxzh on 28/02/2017.
*/
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) {
for (int i = 0; i < (origin.length >> 1); i++) {
int temp = origin[i];
origin[i] = origin[origin.length - 1 - i];
origin[origin.length - 1 - i] = temp;
}
}
/**
* 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 count = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
count++;
}
}
int[] newArray = new int[count];
int j = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 2) {
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 Integer[] merge(int[] array1, int[] array2) {
List<Integer> result = new ArrayList<Integer>();
int i = 0, j = 0;
while (i < array1.length && j < array2.length) {
if (i != array1.length - 1 && array1[i] == array1[i + 1]) {
i++;
continue;
}
if (j != array2.length - 1 && array2[j] == array2[j + 1]) {
j++;
continue;
}
if (array1[i] > array2[j]) {
result.add(array2[j++]);
} else if (array1[i] < array2[j]) {
result.add(array1[i++]);
} else if (array1[i] == array2[j]) {
result.add(array1[i]);
i++;
j++;
}
}
while (i < array1.length) {
result.add(array1[i++]);
}
while (j < array2.length) {
result.add(array2[j++]);
}
return result.toArray(new Integer[result.size()]);
}
/**
* 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 Integer[] fibonacci(int max) {
if (max <= 1)
return new Integer[0];
List<Integer> list = new ArrayList<Integer>();
list.add(1);
list.add(1);
int last = 1;
int count = list.size();
while (last < max) {
int x = list.get(count - 1) + list.get(count - 2);
list.add(x);
count++;
last = x;
}
list.remove(count - 1);
return list.toArray(new Integer[list.size()]);
}
/**
* 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 Integer[] getPrimes(int max) {
List<Integer> list = new ArrayList<Integer>();
if (max < 2)
return new Integer[0];
else if (max == 2)
return new Integer[]{2};
else {
list.add(2);
for (int i = 3; i <= max; i += 2) {
boolean flag = true;
for (int j = 2; j < i; j++) {
if (i % j == 0) {
flag = false;
}
}
if (flag) {
list.add(i);
}
}
if (list.get(list.size() - 1) >= max)
list.remove(list.size() - 1);
}
return list.toArray(new Integer[list.size()]);
}
/**
* 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 Integer[] getPerfectNumbers(int max) {
List<Integer> list = new ArrayList<Integer>();
for (int i = 1; i <= max; i++) {
int sum = 0;
for (int j = 1; j < i; j++) {
if (i % j == 0) {
sum += j;
}
}
if (sum == i) {
System.out.println(i);
list.add(i);
}
}
return list.toArray(new Integer[list.size()]);
}
/**
* 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 seperator
* @return
*/
public String join(int[] array, String seperator) {
StringBuilder builder = new StringBuilder();
for (int i = 0; i < array.length; i++) {
if (i == array.length - 1)
builder.append(array[i]);
else
builder.append(array[i]).append(seperator);
}
return builder.toString();
}
}