/**
* 问题点: 没写注释,代码比较难读。尤其 merge方法。
*/
package com.github.eulerlcs.jmr.algorithm;
import java.util.Arrays;
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 static void reverseArray(int[] origin) {
if (origin == null || origin.length < 2) {
return;
}
for (int head = 0, tail = origin.length - 1; head < tail; head++, tail--) {
origin[head] = origin[head] ^ origin[tail];
origin[tail] = origin[head] ^ origin[tail];
origin[head] = origin[head] ^ origin[tail];
}
}
/**
* 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) {
if (oldArray == null) {
return new int[0];
}
int count = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0)
count++;
}
int[] newArray = new int[count];
int newIndex = 0;
for (int i = 0; i < oldArray.length; i++) {
if (oldArray[i] != 0) {
newArray[newIndex] = oldArray[i];
newIndex++;
}
}
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 static int[] merge(int[] array1, int[] array2) {
if (array1 == null || array1.length == 0) {
if (array2 == null || array2.length == 0) {
return new int[0];
} else {
return Arrays.copyOf(array2, array2.length);
}
} else if (array2 == null || array2.length == 0) {
return Arrays.copyOf(array1, array1.length);
}
int[] result = new int[array1.length + array2.length];
int idxResult = 0;
int idx1 = 0;
int idx2 = 0;
for (; idx1 < array1.length; idx1++) {
if (array1[idx1] < array2[idx2]) {
result[idxResult] = array1[idx1];
idxResult++;
} else if (array1[idx1] == array2[idx2]) {
result[idxResult] = array1[idx1];
idxResult++;
idx2++;
} else {
for (; idx2 < array2.length; idx2++) {
if (array2[idx2] < array1[idx1]) {
result[idxResult] = array2[idx2];
idxResult++;
} else {
if (array2[idx2] == array1[idx1]) {
idx2++;
}
break;
}
}
if (idx2 == array2.length) {
break;
} else {
idx1--;
}
}
}
if (idx1 < array1.length) {
System.arraycopy(array1, idx1, result, idxResult, array1.length - idx1);
idxResult += array1.length - idx1;
}
if (idx2 < array2.length) {
System.arraycopy(array2, idx2, result, idxResult, array2.length - idx2);
idxResult += array2.length - idx2;
}
result = removeZero(result);
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 increaseCapacity
* @return
*/
public static int[] grow(int[] oldArray, int increaseCapacity) {
if (oldArray == null || increaseCapacity < 0) {
return new int[0];
}
int newCapacity = oldArray.length + increaseCapacity;
int[] newArray = Arrays.copyOf(oldArray, newCapacity);
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 static int[] fibonacci(int max) {
if (max <= 1) {
return new int[0];
}
int[] result = new int[10];
result[0] = 1;
result[1] = 1;
int idx = 2;
int sum = 2;
while (sum < max) {
if (idx >= result.length) {
grow(result, result.length * 2);
}
result[idx] = sum;
sum = result[idx - 1] + result[idx];
idx++;
}
result = removeZero(result);
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[2,3,5,7,11,13,17,19]
*
* @param max
* @return
*/
public static int[] getPrimes(int max) {
if (max < 2) {
return new int[0];
}
int[] all = new int[max];
int index = 0;
int temp = 0;
for (int i = 0; i < max; i++) {
all[i] = i;
}
all[0] = 0;
all[1] = 0;
index = 2;
// 筛法
loops: for (; index < max;) {
for (int i = 2;; i++) {
temp = index * i;
if (temp >= max) {
break;
}
all[temp] = 0;
}
for (int i = index + 1; i < max; i++) {
if (all[i] != 0) {
index = i;
continue loops;
}
}
break;
}
int[] result = removeZero(all);
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
*/
public static long[] getPerfectNumbers(long max) {
long[] perfect = new long[49];// 到2016年1月为止,共发现了49个完全数
int idx = 0;
long sum = 1;
long sqrt = 0;
for (long n = 2; n < max; n++) {
sum = 1;
sqrt = (long) Math.sqrt(n);
for (long i = 2; i <= sqrt; i++) {
if (n % i == 0)
sum += i + n / i;
}
if (sum == n) {
perfect[idx] = n;
idx++;
}
}
// return removeZero(perfect);
return perfect;
}
/**
* 用separator 把数组 array给连接起来 例如array= [3,8,9], separator = "-" 则返回值为"3-8-9"
*
* @param array
* @param separator
* @return
*/
public static String join(int[] array, String separator) {
if (array == null || array.length == 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < array.length - 1; i++) {
sb.append(array[i] + separator);
}
sb.append(String.valueOf(array[array.length - 1]));
return sb.toString();
}
}