package org.coding.two.array;
/**
* 数组定长:考虑长度问题;排序之后的数组操作起来更方便;
* @author Administrator
*
*/
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){
if(origin == null || origin.length < 2) {
return;
}
int headIndex = 0;
int lastIndex = origin.length - 1;
while (headIndex < lastIndex) {
int temp = origin[headIndex];
origin[headIndex] = origin[lastIndex];
origin[lastIndex] = temp;
headIndex++;
lastIndex--;
}
}
/**
* 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){
if(oldArray == null || oldArray.length == 0) {
return new int[]{};
}
// int[] newArray = removeZero1(oldArray);
int[] newArray = removeZero2(oldArray);
return newArray;
}
private int[] removeZero2(int[] oldArray) {
int[] indexArray = new int [oldArray.length];
int index = 0;
for(int i = 0, size = oldArray.length; i < size ; i++) {
if(oldArray[i] != 0) {
indexArray[index] = oldArray[i];
index++;
}
}
if(index == 0) {
return new int[]{};
}
int[] newArray = new int[index];
System.arraycopy(indexArray, 0, newArray, 0, index);
return newArray;
}
private int[] removeZero1(int[] oldArray) {
int length = 0;
for(int i = 0, size = oldArray.length; i < size ; i++) {
if(oldArray[i] != 0) {
length++;
}
}
if(length == 0) {
return new int[]{};
}
int[] newArray = new int[length];
for(int i = 0, size = oldArray.length, index = 0; i < size ; i++) {
if(oldArray[i] != 0) {
newArray[index] = oldArray[i];
index++;
}
}
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){
boolean flag1 = (array1 == null || array1.length == 0);
boolean flag2 = (array2 == null || array2.length == 0);
if(flag1 && ! flag2) { //array1 为空
return array2;
}
if(flag2 && !flag1) { //array2 为空
return array1;
}
if(flag1 && flag2){ //array1 和 array2 为空
return new int[0];
}
// array1 和 array2 都不为空
int length1 = array1.length;
int length2 = array2.length;
int index1 = 0;
int index2 = 0;
int newLength = length1 + length2;
int[] newArray = new int[newLength];
int newIndex = 0;
while (index1 < length1 && index2 < length2) {
int val1 = array1[index1];
int val2 = array2[index2];
if(val1 < val2) { //小于
newArray[newIndex] = val1;
newIndex++;
index1++;
} else if(val1 == val2) { //等于
newArray[newIndex] = val1;
newIndex++;
index1++;
index2++;
} else { //大于
newArray[newIndex] = val2;
newIndex++;
index2++;
}
}
//剩余的
while(index1 < length1) {
newArray[newIndex] = array1[index1];
newIndex++;
index1++;
}
while(index2 < length2) {
newArray[newIndex] = array2[index1];
newIndex++;
index2++;
}
//
if(newIndex == newLength) {
return newArray;
}
int[] resutlArray = new int[newIndex];
System.arraycopy(newArray, 0, resutlArray, 0, newIndex);
return resutlArray;
}
/**
* 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){
boolean flag = (oldArray == null || oldArray.length == 0);
boolean flag2 = (size == 0);
if(flag) {
return new int[0];
}
if(flag2 && !flag) {
return oldArray;
}
int length = oldArray.length;
int[] newArray = new int[length + size];
System.arraycopy(oldArray, 0, newArray, 0, 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 < 2) {
return new int[0];
}
// return fibonacci1(max);
return fibonacci2(new int[]{1, 1}, max);
}
/**
* 递归
* @param array
* @param max
* @return
*/
private int[] fibonacci2(int[] array, int max) {
int length = array.length;
if((array[length - 1] + array[length -2]) >= max) {
return array;
}
array = grow(array, 1);
length = array.length;
array[ length - 1] = array[length -2] + array[length -3];
return fibonacci2(array, max);
}
/**
* 循环
* @param max
* @return
*/
private int[] fibonacci1(int max) {
int[] array = new int[]{1, 1};
int length = array.length;
while((array[length - 1] + array[length -2]) < max) {
array = grow(array, 1);
length = array.length;
array[ length - 1] = array[length -2] + array[length -3];
}
return array;
}
/**
* 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){
if(max < 3) {
return new int[0];
}
int[] array = new int[]{2};
int length = array.length;
int val = array[length - 1] + 1;
while(val < max) {
if(isPrime(val)){
array = grow(array, 1);
length++;
array[length - 1] = val;
}
val++;
}
return array;
}
private boolean isPrime(int val) {
for(int i = 2; i < val; i++) {
if(val % i == 0) {
return false;
}
}
return true;
}
/**
* 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;
}
/**
* 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){
if(array == null) {
return null;
}
if(array.length == 0) {
return "";
}
if(seperator == null) {
seperator = "";
}
StringBuilder sb = new StringBuilder();
for (int i : array) {
sb.append(i).append(seperator);
}
return sb.substring(0, sb.length() - seperator.length());
}
}