package com.coding.basic.array;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
public class ArrayUtil {
private final Logger logger = LoggerFactory.getLogger(this.getClass());
/**
* 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 = origin.length;
int mid = length >> 1;
for (int i = 0; i < mid; i++) {
int hIndex = length - 1 -i;
int l = origin[i];
int h = origin[hIndex];
origin[hIndex] = l;
origin[i] = h;
}
}
/**
* 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 removeValue = 0;
return removeValue(oldArray, removeValue);
}
private int[] removeValue(int[] oldArray, int removeValue) {
int length = oldArray.length;
int[] dest = new int[length];
int j = 0;
for(int i = 0; i < length; i++) {
int v = oldArray[i];
if (v != removeValue) {
dest[j++] = v;
}
}
int[] retArray = new int[j];
System.arraycopy(dest, 0, retArray, 0, j);
return retArray;
}
/**
* 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
* todo 数组 a1, b1 自身去重
* @param array1
* @param array2
* @return
*/
public int[] merge(int[] array1, int[] array2){
int length1 = array1.length;
int length2 = array2.length;
int length = length1 + length2;
int[] newArray = new int[length];
return findAndSetLeastWithOutDuplicate(array1, array2, 0, 0, 0, 0, newArray);
}
/**
* todo 优化递归出口判断, 优化三个条件判断为一个
* @param array1
* @param array2
* @param i
* @param j
* @param k
* @param duplicate
* @param newArray
* @return
*/
private int[] findAndSetLeastWithOutDuplicate(int[] array1, int[] array2, int i, int j, int k, int duplicate, int[] newArray) {
if (i == array1.length && j < array2.length) {
System.arraycopy(array2, j, newArray, k, array2.length - j);
return copyLastValues(newArray, duplicate);
}
if (j == array2.length && i < array1.length) {
System.arraycopy(array1, i, newArray, k, array1.length - i);
return copyLastValues(newArray, duplicate);
}
if (j == array2.length && i == array1.length) {
return copyLastValues(newArray, duplicate);
}
int v1 = array1[i];
int v2 = array2[j];
if (v1 < v2) {
newArray [k] = v1;
return findAndSetLeastWithOutDuplicate(array1, array2, ++i, j, ++k, duplicate, newArray);
} else if (v1 > v2){
newArray [k] = v2;
return findAndSetLeastWithOutDuplicate(array1, array2, i, ++j, ++k, duplicate, newArray);
} else {
newArray [k] = v2;
return findAndSetLeastWithOutDuplicate(array1, array2, ++i, ++j, ++k, ++duplicate, newArray);
}
}
private int[] copyLastValues(int[] newArray, int duplicate) {
int[] retArray = new int[newArray.length - duplicate];
System.arraycopy(newArray, 0, retArray, 0, retArray.length);
return retArray;
}
/**
* 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];
System.arraycopy(oldArray, 0, newArray, 0, oldArray.length);
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 int[] fibonacci(int max){
if (max <= 1) {
return new int[]{};
}
int [] newArray = new int[max];
newArray[0] = 1;
return fibonacciN(1, 1, 1, max, newArray);
}
private int[] fibonacciN(int size, int current, int next, int max, int[] newArray) {
if (next >= max) {
int[] retArray = new int[size];
System.arraycopy(newArray, 0, retArray, 0, size);
return retArray;
} else {
newArray[++size - 1] = next;
return fibonacciN(size, next, current + next, max, newArray);
}
}
/**
* 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]
* todo 使用已有的质数序列 优化质数验证
* @param max
* @return
*/
public int[] getPrimes(int max){
if (max <= 2) {
return new int[]{};
}
int[] newArray = new int[max];
int j = 0;
for (int i = 2; i < max; i++) {
if (isPrime(i)) {
newArray[j++] = i;
}
}
int[] retArray = new int[j];
System.arraycopy(newArray, 0, retArray, 0, j);
return retArray;
}
private boolean isPrime(int number) {
int limit = Double.valueOf(Math.sqrt(number)).intValue();
for(int i = 2; i <= limit; i++) {
if (number % i == 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 int[] getPerfectNumbers(int max){
int[] newArray = new int[48]; // 经过不少数学家研究,到2013年2月6日为止,一共找到了48个完全数。
int j = 0;
for (int i = 2; i < max; i++) {
if (isPerfectNumber(i)) {
if (j >= newArray.length) {
newArray = this.grow(newArray, 1);
}
newArray[j++] = i;
}
}
int[] retArray = new int[j];
System.arraycopy(newArray, 0, retArray, 0, j);
return retArray;
}
private boolean isPerfectNumber(int number) {
int sum = 0;
for (int i = 1; i < number; i++) {
if (number % i == 0) {
sum += i;
}
}
return sum == number;
}
/**
* 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(20);
int length = array.length;
for (int i = 0; i< length; i++) {
builder.append(array[i]).append(seperator);
}
builder.deleteCharAt(builder.length() - seperator.length());
return builder.toString();
}
}