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jiangshaojie committed Mar 4, 2017
1 parent f8f3656 commit 1ba1b85
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283 changes: 283 additions & 0 deletions group15/1502_1617273078/src/com/coderising/array/ArrayUtil.java
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package com.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[] a = new int[origin.length];
for (int i = 0; i < origin.length; i++) {
a[i] = origin[origin.length - 1 - i];
}
for (int i = 0; i < a.length; i++) {
System.out.print(a[i]);

}

}

/**
* 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 count = 0;
int index = 0;
//int[] brige = new int[oldArray.length];
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {

count++;
}
}
int[] result = new int[count];
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
result[index++] = oldArray[i];
}


}
return result;
}

/**
* 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 alength = array1.length;
int blength = array2.length;
int[] newint = new int[alength + blength];
for (int i = 0; i < alength; i++) {
newint[i] = array1[i];
}
int index = alength;
//有相同项为true,没有为false
boolean flag = false;
for (int c = 0; c < blength; c++) {
for (int j = 0; j < alength; j++) {
if (array1[j] == array2[c]) {

flag = true;
break;
}

}
if (flag) {

flag = false;
} else {
newint[index] = array2[c];
index++;
}

}
// 去零
newint = removeZero(newint);
//排序

quickSort(newint);
return newint;
}

/**
* 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[] newarry = new int[oldArray.length + size];
for (int i = 0; i < oldArray.length; i++) {
newarry[i] = oldArray[i];
}
return newarry;
}

/**
* 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 count = 0;
for (int i = 0; ; i++) {
if (createfibonacci(i + 1) < max) {
count++;
} else {
break;

}
}
int[] arry = new int[count];
for (int a = 0; a < count; a++) {
arry[a] = createfibonacci(a + 1);
}
return arry;
}

/**
* 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) {
int count = 0;
for (int i = 0; i < max; i++) {
if (isprime(i)) {
count++;
}
}
int[] arry = new int[count];
int sign = 0;
for (int i = 0; i < max; i++) {
if (isprime(i)) {
arry[sign] = i;
sign++;
}
}
return arry;
}

/**
* 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 count = 0;
for (int i = 0; i < max; i++) {
if (isperfectnmber(i)) {
count++;
}
}
int[] arry = new int[count];
int sign = 0;
for (int i = 0; i < max; i++) {
if (isperfectnmber(i)) {
arry[sign] = i;
sign++;
}
}
return arry;

}

/**
* 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
* @return
*/
public String join(int[] array, String seperator) {
String stringBuilder=new String(String.valueOf(array[0]));
for (int i = 1; i <array.length ; i++) {
stringBuilder = stringBuilder + seperator + new String(String.valueOf(array[i]));

}
return stringBuilder;
}

//快排
public static void quickSort(int[] arr) {
qsort(arr, 0, arr.length - 1);
}

private static void qsort(int[] arr, int low, int high) {
if (low < high) {
int pivot = partition(arr, low, high); //将数组分为两部分
qsort(arr, low, pivot - 1); //递归排序左子数组
qsort(arr, pivot + 1, high); //递归排序右子数组
}
}

private static int partition(int[] arr, int low, int high) {
int pivot = arr[low]; //枢轴记录
while (low < high) {
while (low < high && arr[high] >= pivot) --high;
arr[low] = arr[high]; //交换比枢轴小的记录到左端
while (low < high && arr[low] <= pivot) ++low;
arr[high] = arr[low]; //交换比枢轴小的记录到右端
}
//扫描完成,枢轴到位
arr[low] = pivot;
//返回的是枢轴的位置
return low;
}

//生成斐波那契数列
public static int createfibonacci(int n) {
if (n <= 2) {
return 1;
} else {
return createfibonacci(n - 1) + createfibonacci(n - 2);
}
}

//判断是否是素数
public static boolean isprime(int a) {
boolean flag = true;
if (a < 2) {
return false;
} else {
for (int i = 2; i <= Math.sqrt(a); i++) {
if (a % i == 0) {
flag = false;
break;
}
}
}
return flag;
}

//判断是否是完数
public static boolean isperfectnmber(int a) {
boolean flag = true;
int temp = 0;// 定义因子之和变量

for (int n = 1; n < a / 2 + 1; n++) {
if (a % n == 0) {
temp += n;// 能被整除的除数则被加到temp中
}
}
if (temp == a) {// 如果因子之和与原数相等的话,说明是完数
//System.out.print(a + " ");// 输出完数
flag = true;
} else {
flag = false;
}
return flag;
}

}
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