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Merge pull request #20 from lenovo2512009/master
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第三次作业提交
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dracome authored Mar 31, 2017
2 parents 7fff0f5 + 794fbb2 commit b52e86a
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345 changes: 345 additions & 0 deletions group09/610673813/src/coding/week02/array/ArrayUtil.java
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package coding.week02.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)
{
for(int i=0, j = origin.length-1; i<j; i++, j--)
{
int t = origin[i];
origin[i] = origin[j];
origin[j] = t;
}
}

/**
* 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)
{ //传进空数组是返回空数组
if(oldArray == null)
return null;
int count = 0; //统计非零元素个数
int b[] = new int[oldArray.length];
//先统计非零元素个数,并将非零元素存入一个和原数组同样大小的新数组
for(int i=0; i < oldArray.length; i++)
{
if(oldArray[i] != 0)
{
b[count++] = oldArray[i];
}
}
//初始化一个元素个数为非零元素个数的新数组
int newArray[] = new int[count];
//将非零元素copy到新数组
System.arraycopy(b, 0, newArray, 0, count);

/*
* int k=0;
for(int i=0; i<oldArray.length; i++)
{
if(oldArray[i] != 0)
{
newArray[k++] = 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)
{ //当array1和array2都为空时,返回空
if(array1 == null && array2 == null)
return null;
int[] newArray = new int[array1.length + array2.length];
//应该让a1,a2两个数组先进行比较 比较后插入元素
int i = 0; //array1下标
int j = 0; //array2下标
int count = 0; //array3下标
while(i < array1.length && j < array2.length)
{
if(array1[i] < array2[j])
{
newArray[count++] = array1[i++];
}
else if(array1[i] > array2[j])
{
newArray[count++] = array2[j++];
}
else if(array1[i] == array2[j])
{
newArray[count++] = array2[j++];
i++;
}
}
while(i==array1.length && j<array2.length)
{
newArray[count++] = array2[j++];
}
while(j==array2.length && i<array1.length)
{
newArray[count++] = array1[i++];
}

int[] newArray1 = new int[count];
System.arraycopy(newArray, 0, newArray1, 0, count);

return newArray1;
}
/**
* 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
* @throws Exception
*/
public int[] grow(int [] oldArray, int size)
{
if(oldArray == null)
return null;
if(size < 0)
throw new IndexOutOfBoundsException("size小于0");
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[0];
//先将max设置为数组长度,但会浪费空间
int[] a = new int[max];
a[0] = 1;
a[1] = 1;
for(int i=2; i<max; i++)
a[i] = a[i-1] + a[i-2];
//再将max与数组中元素进行比较,获得元素节点位置
int j = 0;
for(j = 0; j<a.length; j++)
{
if(max < a[j])
break;
}
int[] newArray = new int[j];
System.arraycopy(a, 0, newArray, 0, j);

return 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]
* @param max
* @return
*/
public int[] getPrimes(int max)
{
/*
* 思路:先生成素数数组,数组的最大值小于max
*/
//max小于3时,返回空数组
if(max < 3)
return new int[0];
int[] Array = new int[max];
int count = 0; //
int n = 0;
//判断小于max的数有哪些是素数
for(n = 2; n < max; n++)
{
if( count < max)
{
//判断当前n是不是素数
int i = 2;
while(i < n)
{
if(n % i == 0)
break;
if(n % i != 0)
i++;
}
if(i == n)
{ //将素数统计出来
Array[count++] = n;
}
}
}
int[] newArray = new int[count];
System.arraycopy(Array, 0, newArray, 0, count);
return newArray;
}

/**
* 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[] Array = new int[max];

int n = 0;
int count = 0;
int i = 0;
for(n = 2; n < max; n++)
{
int sum = 0;
for(i=1; i<n; i++)
{
if(n%i == 0)
sum += i;
}
if(sum == n)
Array[count++] = n;
}

int[] newArray = new int[count];
System.arraycopy(Array, 0, newArray, 0, count);
return newArray;
}

/**
* 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 s = null;
if(array.length == 0)
return "";
s = Integer.toString(array[0]);
if(array.length > 1)
{
s = s + seperator;
for(int i=1; i<array.length-1; i++)
s = s + array[i] + seperator;
s = s + array[array.length -1];
}
return s;
}


public static void main(String[] args)
{
/* reverseArray()
*
*
int[] a = {1, 2,1, 3, 5, 6};
reverseArray(a);
for(int i=0; i<a.length; i++)
{
System.out.print(a[i] + " ");
}
*/


/* removeZero()
*
*
int oldArr[]={1,3,4,5,0,0,6,6,0,5,4,7,6,7,0,5, 7, 8};
int[] a = removeZero(oldArr);
for(int i=0; i<a.length; i++)
{
System.out.print(a[i] + " ");
}
*/


/* merge()
*
*
int[] a1 = {3, 5, 7, 8, 9};
int[] a2 = {1, 2, 3, 4, 5, 6, 7};
int[] a3 = merge(a1, a2);
for(int i=0; i<a3.length; i++)
{
System.out.print(a3[i] + " ");
}
*/

/* grow()
*
*
int[] a = new int[0];//{3, 5, 7, 8, 9};
int[] newArray = grow(a, -3);
for(int i=0; i<newArray.length; i++)
{
System.out.print(newArray[i] + " ");
}
*/

/* fibonacci()
*
int[] array = fibonacci(35);
for(int i=0; i<array.length; i++)
{
System.out.print(array[i] + " ");
}
*/

/* getPrimes()
*
int[] array = getPrimes(2);
for(int i=0; i<array.length; i++)
{
System.out.print(array[i] + " ");
}
*/

/* getPerfectNumbers()
*
int[] array = getPerfectNumbers(10000);
for(int i=0; i<array.length; i++)
{
System.out.print(array[i] + " ");
}
*/

/* join()
*
int[] Array = {3, 5, 7, 8, 9};
int[] array = {};
String s = join(array, "-");
System.out.println(s);
*/

}


}
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