package net.coding.coderising.array;
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 l = 0, r = origin.length - 1;
while (l < r) {
int tmp = origin[l];
origin[l] = origin[r];
origin[r] = tmp;
l++;
r--;
}
}
/**
* 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 cntZero = 0;
for (int i = 0; i < oldArray.length;i++) {
if (oldArray[i] == 0) {
cntZero ++;
}
}
if (cntZero == 0) {
return oldArray;
}
int[] newArray = new int[oldArray.length - cntZero];
int j = 0;
for (int i = 0; i < oldArray.length;i++) {
if (oldArray[i] != 0) {
newArray[j++] = oldArray[i];
}
}
return newArray;
}
/**
* 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[] result = new int[array1.length + array2.length];
int l = 0;
int r = 0;
int cnt = 0;
while (true) {
if (l >= array1.length && r >= array2.length) {
break;
}
if (l >= array1.length) {
result[cnt++] = array2[r];
r++;
} else if (r >= array1.length) {
result[cnt++] = array1[l];
l++;
} else {
if (array1[l] < array2[r]) {
result[cnt++] = array1[l];
l++;
} else {
result[cnt++] = array2[r];
r++;
}
}
}
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 int[] grow(int [] oldArray, int size){
int[] result = new int[oldArray.length + size];
System.arraycopy(oldArray, 0, result, 0, oldArray.length);
return result;
}
/**
* 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){
int[] result = {};
int a = 0;
int b = 1;
int cnt = 0;
while (b < max) {
result = grow(result, 1);
result[cnt++] = b;
int tmp = a + b;
a = b;
b = tmp;
}
return result;
}
/**
* 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[1,2,3,5,7,11,13,17,19]
* @param max
* @return
*/
private boolean isPrime(int n) {
for (int i = 2; i <= Math.sqrt(n + 1.0); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public int[] getPrimes(int max){
int[] result = {};
int cnt = 0;
for (int i = 1; i < max; i++) {
if (isPrime(i)) {
result = grow(result, 1);
result[cnt++] = i;
}
}
return result;
}
/**
* 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
*/
private boolean isPerfect(int n) {
int total = 0;
for (int i = 1; i < n; i++) {
if (n % i == 0) {
total += i;
}
}
return total == n;
}
public int[] getPerfectNumbers(int max){
int[] result = {};
int cnt = 0;
for (int i = 1; i < max; i++) {
if (isPerfect(i)) {
result = grow(result, 1);
result[cnt++] = i;
}
}
return result;
}
/**
* 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){
StringBuilder sb = new StringBuilder();
for (int i = 0; i < array.length; i++) {
if (sb.length() > 0) {
sb.append(seperator);
}
sb.append(array[i]);
}
return sb.toString();
}
}