package coding;
import java.util.Arrays;
/**
* @author jiaxun
*/
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) return;
int length = origin.length;
for (int i = 0, middle = length / 2; i < middle; i++) {
int temp = origin[i];
origin[i] = origin[length - i - 1];
origin[length - i - 1] = 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) {
if (oldArray == null) return null;
int[] newArray = new int[oldArray.length];
int count = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArray[count++] = oldArray[i];
}
}
return Arrays.copyOf(newArray, count);
}
/**
* 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 len1 = array1.length;
int len2 = array2.length;
int[] array3 = new int[len1 + len2];
int arr1Index = 0, arr2Index = 0, arr3Index = 0, zeroCount = 0;
while (arr1Index < len1 && arr2Index < len2) {
if (array1[arr1Index] < array2[arr2Index]) {
array3[arr3Index++] = array1[arr1Index++];
} else if (array1[arr1Index] > array2[arr2Index]) {
array3[arr3Index++] = array2[arr2Index++];
} else {
array3[arr3Index++] = array1[arr1Index++] = array2[arr2Index++];
zeroCount++;
}
}
while (arr1Index < len1) {
array3[arr3Index++] = array1[arr1Index++];
}
while (arr2Index < len2) {
array3[arr3Index++] = array2[arr2Index++];
}
return Arrays.copyOf(array3, len1 + len2 - zeroCount);
}
/**
* 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){
return Arrays.copyOf(oldArray, oldArray.length + size);
}
/**
* 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){
if (max == 1) return null;
int first = 1, second = 1;
int next = 2;
int[] result = new int[max];
result[0] = first; result[1] = second;
int count = 2;
while (next < max) {
result[count++] = next;
first = second;
second = next;
next = first + second;
}
return Arrays.copyOf(result, count);
}
/**
* 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) return null;
int [] result = new int[max];
int count = 0;
for (int i = 2; i < max; i++) {
if (!primes(i)) {
result[count++] = i;
}
}
return Arrays.copyOf(result, count);
}
private boolean primes(int data) {
int sqrt = (int) Math.sqrt(data) + 1;
for (int i = 2; i < sqrt; i++) {
if (data % i == 0){
return true;
}
}
return false;
}
/**
* 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[max - 2];
int count = 0;
for (int i = 2; i < max; i++) {
if (perfectNumber(i))
result[count++] = i;
}
return Arrays.copyOf(result, count);
}
private boolean perfectNumber(int number) {
int left = 2, right = number;
int sum = 1;
while (left < right) {
if (number % left == 0) {
sum += left;
right = number / left;
if (left != right) sum += right;
}
left++;
}
return sum == number;
}
/**
* 用separator 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 separator
* @return
*/
public String join(int[] array, String separator){
if (array == null) return null;
StringBuilder result = new StringBuilder();
result.append(array[0]);
for (int i = 1, len = array.length; i < len; i++) {
result.append(separator).append(array[i]);
}
return result.toString();
}
}