package com.coderising.array;
import org.junit.experimental.max.MaxCore;
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) {
if (origin.length < 2)
return;
int temp;
for (int i = 0; i < origin.length >> 1; i++) {
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 static int[] removeZero(int[] oldArray) {
if (oldArray == null || oldArray.length == 0)
return null;
int zeros = 0;// 数组中0元素的个数
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] == 0)
zeros++;
}
int[] newArr = new int[oldArray.length - zeros];
for (int i = 0, j = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArr[j] = oldArray[i];
j++;
}
}
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) {
int[] mergedArr = new int[array1.length + array2.length];
int temp;
int i = 0, j = 0, index = 0, size = 0;
while (i < array1.length && j < array2.length) {
//两个数组都没遍历到最后一个元素
if (i != array1.length - 1 && j != array2.length - 1) {
if (array1[i] < array2[j]) {
temp = array1[i++];
} else if (array1[i] > array2[j]) {
temp = array2[j++];
} else {
//遇到相等元素,存放任意一个就实现去重了
temp = array1[i++];
j++;
}
mergedArr[index++] = temp;
size++;
//array1遍历到最后一个元素
} else if (i == array1.length - 1 && j != array2.length - 1) {
if (array1[i] < array2[j]) {
temp = array1[i];
mergedArr[index++] = temp;
size++;
//将array2的剩余元素复制到mergedArr中
System.arraycopy(array2, j, mergedArr, index, array2.length - j);
size += array2.length - j;
break;
} else if (array1[i] > array2[j]) {
temp = array2[j++];
size++;
} else {
System.arraycopy(array2, j, mergedArr, index, array2.length - j);
size += array2.length - j;
break;
}
//array2遍历到最后一个元素
} else if (i != array1.length - 1 && j == array2.length - 1) {
if (array1[i] > array2[j]) {
temp = array2[j];
mergedArr[index++] = temp;
size++;
//将array1的剩余元素复制到mergedArr中
System.arraycopy(array1, i, mergedArr, index, array1.length - i);
size += array1.length - i;
break;
} else if (array1[i] < array2[j]) {
temp = array2[i++];
size++;
} else {
System.arraycopy(array1, i, mergedArr, index, array1.length - i);
size += array1.length - i;
break;
}
}
}
//构造新数组,去除mergedArr中尾部若干0元素
int[] result = new int[size];
System.arraycopy(mergedArr, 0, result, 0, size);
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 static int[] grow(int[] oldArray, int size) {
int[] arr = new int[oldArray.length + size];
System.arraycopy(oldArray, 0, arr, 0, oldArray.length);
return arr;
}
/**
* 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 < 2) return null;
int[] a = new int[max];
a[0] = 1;
a[1] = 1;
int size = 2;
for (int i = 2;; i++) {
a[i] = a[i-1] + a[i-2];
if (a[i] > max) break;
size ++;
}
int[] fibonacci = new int[size];
System.arraycopy(a, 0, fibonacci, 0, size);
return fibonacci;
}
/**
* 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) {
if (max < 2) {
return null;
}
int[] a = new int[max];
int size = 0;
for (int i = 2; i < max; i++) {
if (isPrime(i)) {
a[size++] = i;
}
}
int[] primes = new int[size];
System.arraycopy(a, 0, primes, 0, size);
return primes;
}
private static boolean isPrime(int i) {
for (int j = 2; j*j <= i; j++) {
if (i % j == 0) {
return false;
}
}
return true;
}
/**
* 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) {
int[] a = new int[max];
int size = 0;
for (int i = 6; i < max; i++) {
if (isPerfectNumber(i)) {
a[size++] = i;
}
}
int[] perfectNumbers = new int[size];
System.arraycopy(a, 0, perfectNumbers, 0, size);
return perfectNumbers;
}
private static boolean isPerfectNumber(int i) {
int sum = 0;
for (int j = 1; j <= i >> 1; j++) {
if (i % j == 0) {
sum += j;
}
}
if (i == sum) return true;
else return false;
}
/**
* 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 : array) {
sb.append(i);
sb.append(seperator);
}
return sb.substring(0, sb.lastIndexOf(seperator));
}
}