package com.github.congcongcong250.coding2017.basic;
import java.util.Arrays;
import java.util.ArrayList;
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[] copy = origin.clone();
int j, last = copy.length - 1;
for (int i = last; i >= 0; i--) {
j = last - i;
origin[i] = copy[j];
}
}
/**
* 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) {
/*
* Method 1, traverse oldArray and count non-zero value, create new
* int[count], traverse again oldArray and insert to new one
*/
/*
* Method 2, traverse olArray and insert non-zero value to an List<int>,
* then List.toArray().
*/
// Method 3
if(oldArray == null){
return null;
}
int[] tmp = new int[oldArray.length];
int count = 0;
for (int v : oldArray) {
if (v != 0) {
tmp[count] = v;
count++;
}
}
return Arrays.copyOf(tmp, 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) {
if (array1 == null) {
if (array2 == null) {
return null;
}
return array2;
} else if (array2 == null) {
return array1;
}
int l1 = array1.length, l2 = array2.length;
int[] whole = new int[l1 + l2];
int dupValue = 0;
int dupCount = 0;
boolean isDup = false;
int j = 0;
// Traverse array1
for (int i = 0; i < l1; i++) {
// Get rid of duplicate value in array1
if (isDup) {
if (array1[i] == dupValue) {
dupCount++;
continue;
} else {
isDup = false;
}
}
while (j < l2) {
if (array1[i] > array2[j]) {
// If int from array1 is larger, add int from array2 first
whole[i + j - dupCount] = array2[j];
j++;
continue;
} else if (array1[i] == array2[j]) {
// If equals, skip int from array2, and set duplicate value
isDup = true;
dupValue = array1[i];
dupCount++;
j++;
continue;
} else if (array1[i] < array2[j]) {
// If smaller, break and add from in array1
break;
}
}
whole[i + j - dupCount] = array1[i];
}
// Deal with left over in array2
while (j < l2) {
whole[l1 + j - dupCount] = array2[j];
j++;
}
return Arrays.copyOf(whole, l1 + l2 - dupCount);
}
/**
* 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) {
checkGrowSize(size);
if (oldArray == null) {
return null;
}
int newlength = oldArray.length + size;
int[] res = new int[newlength];
res = Arrays.copyOf(oldArray, newlength);
return res;
}
private void checkGrowSize(int size) {
if (size < 0) {
throw new IndexOutOfBoundsException(
"Negative size is not allowed in grow()");
}
}
/**
* 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[0];
}
// Fib(47) == 2971215073 > INT_MAX;
int[] tmp = new int[46];
tmp[0] = 1;
tmp[1] = 1;
int next = 1 + 1;
int i = 1;
while (next < max) {
i++;
tmp[i] = next;
next = tmp[i] + tmp[i - 1];
}
return Arrays.copyOf(tmp, i + 1);
}
/**
* 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 new int[0];
}
ArrayList<Integer> primeList = new ArrayList<Integer>();
primeList.add(2);
for (int candidate = 3; candidate < max; candidate += 2) {
// For every number smaller than max
int ceiling = (int) Math.floor(Math.sqrt(candidate));
for (Integer prime : primeList) {
// Divided by every prime number smaller than sqrt(candidate)
if (candidate % prime == 0) {
// When reminder equals 0, candidate is not a prime
break;
}
if (prime > ceiling) {
// When every prime number <= sqrt(candidate), add candidate to primelist
primeList.add(candidate);
break;
}
}
}
// Transfer primelist to int array
int[] res = arrayFromList(primeList);
return res;
}
/**
* 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) {
if (max <= 2) {
return new int[0];
}
ArrayList<Integer> perfectList = new ArrayList<Integer>();
for (int candidate = 2; candidate < max; candidate++) {
int sum = 1;
int ceiling = (int) Math.floor(Math.sqrt(candidate));
if (Math.sqrt(candidate) == ceiling) {
sum += ceiling;
}
// For each number smaller than max
for (int divisor = 2; divisor <= ceiling; divisor++) {
if (candidate % divisor == 0) {
sum += divisor;
sum += candidate / divisor;
}
if (sum > candidate) {
break;
}
}
if (sum == candidate) {
perfectList.add(candidate);
}
}
// Transfer primelist to int array
int[] res = arrayFromList(perfectList);
return res;
}
private int[] arrayFromList(ArrayList<Integer> list) {
int[] r = new int[list.size()];
int j = 0;
for (Integer v : list) {
r[j++] = v;
}
return r;
}
/**
* 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) {
String res = "";
if (array == null || array.length == 0) {
return res;
}
res += array[0];
for (int i = 1; i < array.length; i++) {
res += seperator + array[i];
}
return res;
}
}