package com.github.wdn.coding2017.coderising.array;
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 void reverseArray(int[] origin) {
int length = origin.length;
for (int i = 0; i < length/2; i++) {
origin[i]=origin[i]+origin[length-1-i];
origin[length-1-i] = origin[i]-origin[length-1-i];
origin[i] = origin[i]-origin[length-1-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[] result = new int[oldArray.length];
int resultSize=0;
for (int i = 0; i < oldArray.length; i++) {
int item = oldArray[i];
if (item!=0){
result[resultSize]=item;
resultSize++;
}
}
if(resultSize<oldArray.length) {
result = Arrays.copyOfRange(result, 0, resultSize);
}
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){
// 方法一:省事但遍历次数多
int[] mergeArr0 = Arrays.copyOf(array1, array1.length+array2.length);
System.arraycopy(array2,0,mergeArr0,array1.length,array2.length);
Arrays.sort(mergeArr0);
if(mergeArr0.length<2){
return mergeArr0;
}
int[] merge = new int[mergeArr0.length];
merge[0]=mergeArr0[0];
int mergeIndex=0;
// 排重
for (int i = 1; i < mergeArr0.length; i++) {
if(mergeArr0[i]!=merge[mergeIndex]){
mergeIndex++;
merge[mergeIndex]=mergeArr0[i];
}
}
return Arrays.copyOfRange(merge,0,mergeIndex+1);
/*
// 方法二:复杂但遍历次数少
int[] mergeArr = new int[array1.length+array2.length];
int array1Index = 0;
int array2Index = 0;
int mergeArrIndex=0;
for(int i=0;i<mergeArr.length;i++){
if(array1Index>=array1.length && array2Index>=array2.length){
break;
}
int array1Value = array1Index < array1.length ? array1[array1Index] : array2[array2Index];
int array2Value = array2Index < array2.length ? array2[array2Index] : array1[array1Index];
if (array1Value > array2Value) {
mergeArr[i] = array2Value;
array2Index++;
} else if (array1Value < array2Value) {
mergeArr[i] = array1Value;
array1Index++;
}else{
mergeArr[i] = array1Value;
array1Index++;
array2Index++;
}
mergeArrIndex++;
}
return Arrays.copyOfRange(mergeArr,0,mergeArrIndex);
*/
}
/**
* 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){
if (size<0 || oldArray.length+size>Integer.MAX_VALUE){
throw new IndexOutOfBoundsException();
}
// 方法一:使用jdk自带方法
//return Arrays.copyOf(oldArray,oldArray.length+size);
// 方法二:遍历
int[] growArr = new int[oldArray.length+size];
for (int i = 0; i < oldArray.length; i++) {
growArr[i] = oldArray[i];
}
return growArr;
}
/**
* 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 || max>Integer.MAX_VALUE){
throw new IllegalArgumentException();
}
int[] initArr = new int[]{1, 1};
if(max<=2){
return initArr;
}
int aIndex = 0;
int bIndex = 1;
int[] fibonacciArr = Arrays.copyOf(initArr,max);
int index = 2;
while(fibonacciArr[aIndex]+fibonacciArr[bIndex]<max){
fibonacciArr[index]=fibonacciArr[aIndex]+fibonacciArr[bIndex];
index++;
aIndex++;
bIndex++;
}
return Arrays.copyOf(fibonacciArr,index);
}
/**
* 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[] result = new int[max];
int index = 0;
for (int i = 0; i <= max; i++) {
if (isPrimes(i)){
result[index] = i;
index++;
}
}
return Arrays.copyOf(result,index);
}
private boolean isPrimes(int value){
boolean result = true;
if(value<2) return false;
if(value==2) return true;
if(value%2==0) return false;
for (int i = 3; i < value/2; i++) {
if(value%i==0){
result=false;
break;
}
}
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 int[] getPerfectNumbers(int max){
if (max<1){
throw new IllegalArgumentException();
}
// 十万里边就四个,此算法已经算的很慢了
int[] result = new int[4];
int index = 0;
for (int i = 1; i <= max; i++) {
int sum=0;
for (int j = 1; j < i; j++) {
if(i%j==0){
sum+=j;
}
}
if(sum == i){
result[index]=i;
index++;
}
}
return result;
}
/**
* 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){
StringBuffer sb = new StringBuffer();
for (int i = 0; i < array.length; i++) {
if(i<array.length-1){
sb.append(array[i]).append(seperator);
}else{
sb.append(array[i]);
}
}
return sb.toString();
}
}