package com.coderising.array;
import java.util.Arrays;
public final class MyArrayUtil {
public static void main(String[] args) {
int[] reverse1 = {7, 9, 30, 3};
int[] reverse2 = {7, 9, 30, 3, 4};
reverseArray(reverse1);
reverseArray(reverse2);
System.out.println("reverseArray: " + Arrays.toString(reverse1) + " " + Arrays.toString(reverse2));
int[] removeZero = {1, 3, 4, 5, 0, 0, 6, 6, 0, 5, 4, 7, 6, 7, 0, 5};
System.out.println("removeZero: " + Arrays.toString(removeZero(removeZero)));
int[] merge1 = {3, 5, 7, 8};
int[] merge2 = {4, 5, 6, 7};
System.out.println("merge: " + Arrays.toString(merge(merge1, merge2)));
System.out.println("merge: " + Arrays.toString(merge(merge2, merge1)));
int[] grow = {2, 3, 6};
System.out.println("grow: " + Arrays.toString(grow(grow, 3)));
System.out.println("fibonacci: f(15)=" + Arrays.toString(fibonacci(15)) + " f(1)=" + Arrays.toString(fibonacci(1)));
System.out.println("getPrimes: g(23)=" + Arrays.toString(getPrimes(23)));
System.out.println("getPerfectNumbers: g(500)=" + Arrays.toString(getPerfectNumbers(500)));
System.out.println("getPerfectNumbers: j(f(50))=" + join(fibonacci(50), "-"));
}
/**
* 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) {
int len = origin.length;
for (int i = 0; i < len / 2; i++) {
int j = len - 1 - i;
int temp = origin[i];
origin[i] = origin[j];
origin[j] = 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) {
int length = oldArray.length;
// 空间换时间复杂度 O(2*n-m) n为数组长度 m为0的个数
int[] src = new int[length];
int last = 0;
for (int i = 0; i < length; i++) {
int m = oldArray[i];
if (m != 0) {
src[last++] = m;
}
}
int[] dest = new int[last];
System.arraycopy(src, 0, dest, 0, last);
return dest;
}
/**
* 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[] src = new int[array1.length + array2.length];
int i = 0, j = 0, k = 0; //i=数组1游标 j=数组2游标 k=目标数组下标
// 时间复杂度 O(n+m-k) n为数组1长度 m为数组2长度 k为重复数组个数
while (i < array1.length && j < array2.length) {
if (array1[i] < array2[j]) { //大小比较入目标数组
src[k++] = array1[i++];
} else if (array1[i] > array2[j]) {
src[k++] = array2[j++];
} else { //等值同时移动,但只入1个目标值
src[k++] = array1[i++];
j++;
}
}
// 只要1个数组下表越界,就直接添加另1个数组尾部所有有序值
while (i < array1.length) {
src[k++] = array1[i++];
}
while (j < array2.length) {
src[k++] = array2[j++];
}
int[] dest = new int[k];
System.arraycopy(src, 0, dest, 0, k);
return dest;
}
/**
* 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[] newArr = new int[oldArray.length + size];
System.arraycopy(oldArray, 0, newArr, 0, oldArray.length);
return newArr;
}
/**
* 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) { // 额外出口 f(1) = []
return new int[]{};
}
if (max == 2) { //递归出口 f(2) = [1, 1]
return new int[]{1, 1};
}
// f(3) = [1,1,2]
// f(5) = f(4) = [1,1,2,3]
// f(8) = f(7) = f(6) = [1,1,2,3,5]
// 递归拿上个数列最后的两个数
int[] src = fibonacci(max - 1);
// 判断这两个数的和值是否在max范围内
int m = src[src.length - 1] + src[src.length - 2];
if (m < max) {
int[] dest = new int[src.length + 1];
System.arraycopy(src, 0, dest, 0, src.length);
dest[src.length] = m;
return dest;
} else {
return src;
}
}
/**
* 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) { //特殊出口 g(1) = []
return new int[]{};
}
if (max == 3) { //递归出口 g(3) = [2]
return new int[]{2};
}
// g[3] = [2]
// g[5] = g[4] = [2,3]
// g[7] = g[6] = [2,3,5]
// g[11] = g[10] = g[9] = g[8] = [2,3,5,7]
// 递归拿上个数组最后的素数
int[] src = getPrimes(max - 1);
// 找上个素数和最大值之间的素数
int before = src[src.length - 1];
int next = before;
int start = 1 + before;
int end = max;
for (int num = start; num < end; num++) {
int i = 2;
for (; i <= num / 2; i++) {
if (num % i == 0) {
break;
}
}
if (i > num / 2) {
next = num;
break;
}
}
// 判断这个值是否存在
if (next > before) {
int[] dest = new int[src.length + 1];
System.arraycopy(src, 0, dest, 0, src.length);
dest[src.length] = next;
return dest;
} else {
return src;
}
}
/**
* 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) {
if (max <= 6) { //特殊出口
return new int[]{};
}
if (max == 7) { //递归出口
return new int[]{6};
}
// 递归拿上个数组最后的完数
int[] src = getPerfectNumbers(max - 1);
// 找上个完数和最大值之间的完数
int before = src[src.length - 1];
int next = before;
int start = 1 + before;
int end = max;
for (int num = start; num < end; num++) {
int count = 0;
for (int i = 1; i <= num / 2; i++) {
if (num % i == 0) {
count += i;
}
}
if (count == num) {
next = num;
break;
}
}
// 判断这个值是否存在
if (next > before) {
int[] dest = new int[src.length + 1];
System.arraycopy(src, 0, dest, 0, src.length);
dest[src.length] = next;
return dest;
} else {
return src;
}
}
/**
* 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 static String join(int[] array, String seperator) {
StringBuilder sBuilder = new StringBuilder();
for (int i = 0; i < array.length; i++) {
sBuilder.append(array[i]).append(seperator);
}
if (sBuilder.length() > 0) {
sBuilder.deleteCharAt(sBuilder.length() - 1);
}
return sBuilder.toString();
}
}