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Merge pull request #27 from duxiaolong/master
完成ArrayUtil编码
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group16/1287642108/0305/src/com/coderising/array/ArrayUtil.java
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package com.coderising.array; | ||
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import java.util.ArrayList; | ||
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public class ArrayUtil { | ||
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/** | ||
* 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] | ||
* 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){ | ||
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public void reverseArray(int[] origin) { | ||
int[] temp = new int[origin.length]; | ||
int index = 0; | ||
for (int i = origin.length - 1; i >= 0; i--) { | ||
temp[index++] = origin[i]; | ||
} | ||
} | ||
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/** | ||
* 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} | ||
* 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 | ||
*/ | ||
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public int[] removeZero(int[] oldArray){ | ||
return null; | ||
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public static int[] removeZero(int[] oldArray) { | ||
int countZero = 0; | ||
for (int i = 0; i < oldArray.length; i++) { | ||
if(oldArray[i] == 0){ | ||
countZero++; | ||
} | ||
} | ||
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int[] temp = new int[oldArray.length - countZero]; | ||
int index = 0; | ||
for (int i = 0; i < oldArray.length; i++) { | ||
if (oldArray[i] != 0) { | ||
temp[index++] = oldArray[i]; | ||
} | ||
} | ||
return temp; | ||
} | ||
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/** | ||
* 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 | ||
* 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 | ||
*/ | ||
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public int[] merge(int[] array1, int[] array2){ | ||
return null; | ||
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public Integer[] merge(int[] array1, int[] array2) { | ||
Integer[] temp = new Integer[array1.length + array2.length]; | ||
int i = 0,j = 0; | ||
int index = 0; | ||
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while(i < array1.length && j < array2.length){ | ||
if(array1[i] <= array2[j]){ | ||
temp[index++] = array1[i++]; | ||
}else{ | ||
temp[index++] = array2[j++]; | ||
} | ||
} | ||
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while(i < array1.length){ | ||
temp[index++] = array1[i++]; | ||
} | ||
while(j < array2.length){ | ||
temp[index++] = array2[j++]; | ||
} | ||
return removeRepetition(temp); | ||
} | ||
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/** | ||
* 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] | ||
* 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){ | ||
return null; | ||
public static int[] grow(int[] oldArray, int size) { | ||
int[] temp = new int[oldArray.length + size]; | ||
System.arraycopy(oldArray, 0, temp, 0, oldArray.length); | ||
return temp; | ||
} | ||
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/** | ||
* 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 [] | ||
* 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){ | ||
return null; | ||
public static int[] fibonacci(int max){ | ||
int[] temp = new int[max]; | ||
if(max == 1){ | ||
return null; | ||
} | ||
int index = 0; | ||
for(int i = 0;i<max;i++){ | ||
if(fibonacciNum(i) < max){ | ||
temp[index++] = fibonacciNum(i); | ||
} | ||
} | ||
return removeZero(temp); | ||
} | ||
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public static int fibonacciNum(int n) { | ||
if (n <= 1) { | ||
return 1; | ||
} | ||
return fibonacciNum(n-1) + fibonacciNum(n-2); | ||
} | ||
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/** | ||
* 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] | ||
* 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){ | ||
return null; | ||
public static int[] getPrimes(int max){ | ||
boolean isPrime = false; | ||
ArrayList<Integer> temp = new ArrayList<>(); | ||
for(int i=2; i < max; i++){ | ||
isPrime = true; | ||
for (int j = 2; j < i; j++){ | ||
if ((i % j) == 0) { | ||
isPrime = false; | ||
break; | ||
} | ||
} | ||
if(isPrime){ | ||
temp.add(i); | ||
} | ||
} | ||
int[] array = new int[temp.size()]; | ||
int index = 0; | ||
for(Integer te : temp){ | ||
array[index++] = te; | ||
} | ||
return array; | ||
} | ||
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/** | ||
* 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 | ||
* 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){ | ||
return null; | ||
public static Integer[] getPerfectNumbers(int max) { | ||
Integer[] temp = new Integer[max]; | ||
int index = 0; | ||
for (int i = 2; i < max; i++) { | ||
int sum = 0; | ||
for (int j = 1; j < i; j++) { | ||
if (i % j == 0) { | ||
sum += j; | ||
} | ||
if (sum == i){ | ||
temp[index++] = i; | ||
} | ||
} | ||
} | ||
return removeRepetition(temp); | ||
} | ||
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/** | ||
* 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" | ||
* 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){ | ||
return null; | ||
public String join(int[] array, String seperator) { | ||
String temp = ""; | ||
for(int i = 0;i < array.length -1; i++){ | ||
temp += array[i] + seperator; | ||
} | ||
temp += array[array.length-1]; | ||
return temp; | ||
} | ||
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public static Integer[] removeRepetition(Integer[] oldArray){ | ||
ArrayList<Integer> temp = new ArrayList<>(); | ||
for(int i=0;i<oldArray.length;i++){ | ||
for(int j=i+1;j<oldArray.length;j++){ | ||
if(oldArray[i] == oldArray[j]){ | ||
j = ++i; | ||
} | ||
} | ||
temp.add(oldArray[i]); | ||
} | ||
Integer[] result = (Integer[]) temp.toArray(new Integer[temp.size()]); | ||
return result; | ||
} | ||
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} |