[英]Displaying odd values in an array
我試圖在數組中顯示奇數,但每個數字僅顯示一次(即,numbers [3] = 3,3,1;僅顯示3和1而不是3、3和1。)
這是我到目前為止的代碼,該程序將完全創建一個具有用戶輸入的特定長度的,然后將計算數組中的最大最小值和奇數值。
import java.util.Scanner;
public class ArrayLab
{
static Scanner input = new Scanner (System.in);
public static void main(String[] args)
{
System.out.println("Enter the number of numbers: ");
final int NUMBER_OF_ELEMENTS = input.nextInt();
double[] numbers = new double[NUMBER_OF_ELEMENTS];
System.out.println("Enter the numbers: ");
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++)
{
numbers[i] = input.nextDouble();
}
input.close();
double max = numbers[0];
double min = numbers[0];
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++)
{
if (numbers[i] > max)
{
max = numbers[i];
}
}
System.out.println("The max is: " + max);
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++)
{
if (numbers[i] < min)
{
min = numbers[i];
}
}
System.out.println("The min is: " + min);
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++)
{
if (numbers[i] % 2 != 0)
{
System.out.println ("The odd numbers are: " + numbers[i]);
}
}
}
}
謝謝你的幫助。
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++)
{
if (numbers[i] % 2 != 0)
{
set.add(numbers[i]);
}
}
System.out.println ("The odd numbers are: " +set);
使用Java8可以簡化很多工作:
double[] d = Arrays.toStream(numbers).filter(d -> (d % 2) == 1).distinct().toArray();
for(double tmp : d)
System.out.println(tmp);
System.out.println("min: " + Arrays.toStream(numbers).min((a , b) -> new Double(a).compareTo(b)));
System.out.println("max: " + Arrays.toStream(numbers).max((a , b) -> (new Double(a).compareTo(b))));
對於您的解決方案:您永遠不會消除重復的數字,因此重復項會保留在數組中,直到您打印所有的奇數和最大數為止。
可以通過以下幾種方法消除這種情況:
Set
,因為這些值不允許重復值 根據您的需要更新了解決方案。 並且請使用更好的編碼標准。 請注意,條件檢查!oddNumbers.contains(numbers [i])並不是非常必要,因為HashSet絕不使用任何重復的值。
import java.util.HashSet;
import java.util.Scanner;
public class ArrayLab {
static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
System.out.println("Enter the number of numbers: ");
final int NUMBER_OF_ELEMENTS = input.nextInt();
double[] numbers = new double[NUMBER_OF_ELEMENTS];
System.out.println("Enter the numbers: ");
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++) {
numbers[i] = input.nextDouble();
}
input.close();
HashSet<Double> oddNumbers = new HashSet<Double>(NUMBER_OF_ELEMENTS);
double max = numbers[0];
double min = numbers[0];
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++) {
if (numbers[i] > max) {
max = numbers[i];
}
if (numbers[i] < min) {
min = numbers[i];
}
if (numbers[i] % 2 != 0 && !oddNumbers.contains(numbers[i])) {
oddNumbers.add(numbers[i]);
}
}
System.out.println("The max is: " + max);
System.out.println("The min is: " + min);
System.out.println("The odd numbers are: " + oddNumbers);
}
}
對於您的方法,更有意義的解決方案如下:
int[] tempArray; //temporary array to store values from your original "array"
int count=0;
for(int i=0; i<numbers.length; i++) {
if(numbers[i]%2 != 0) {
count++;
}
}
tempArray = new int[count]; //initializing array of size equals to number of odd digits in your array
int j = 0;
for(int i=0; i<numbers.length; i++) {
boolean check = true;
for(int k=0; k<j; k++) {
if(tempArray[k] == numbers[i]) {
check = false; //this will prevent duplication of odd numbers
}
}
if(numbers[i]%2 != 0 && check) {
tempArray[j]=numbers[i];
j++;
}
}
//Now print the tempArray which contains all the odd numbers without repetition
少數人提到了集合,但是還有另一種方式。 只需對數組進行排序,然后進行掃描,就最后一個打印出的數字進行檢查。 即
int lastPrinted = 0;
// Sort the array
Arrays.sort(numbers);
System.out.print("The odd numbers are: ");
// Scan through the array
for (int i = 0; i < NUMBER_OF_ELEMENTS; i++)
{
// if it's odd and doesn't match the last one...
if (numbers[i] % 2 != 0 && numbers[i] != lastPrinted)
{
// ...print it and update lastPrinted
System.out.print( "" + numbers[i] );
lastPrinted = numbers[i];
}
}
System.out.println("");
附帶說明一下,您實際上不必遍歷整個陣列兩次即可找到最大值和最小值,您可以一口氣做到這一點。
我認為您可以使用內置的hashmap類及其方法來完成任務,而不會在很大程度上影響算法的復雜性。
import java.util.HashMap;
公共類散列{
public static void main(String[] args) {
//declare a new hasmap
HashMap<Integer, Integer> map = new HashMap<>();
//consider Arr as your Array
int Arr[] = {3,3,1,4,5,5,7,8};
//traverse through the array
for(int i=0;i<Arr.length;i++){
//check if the required condition is true
if(Arr[i]%2==1){
/*now we insert the elements in the map but before
that we have to make sure that we don't insert duplicate values*/
if(!map.containsKey(Arr[i])){// this would not affect the complexity of Algorithm since we are using hashMap
map.put(Arr[i], Arr[i]);//We are storing the Element as KEY and as VALUE in the map
}
}
}
//now We can get these element back from map by using the following statement
Integer[] newArray = map.values().toArray(new Integer[0]);
//All the required elements are now present in newArray
for(int ele:newArray){
System.out.println(ele);
}
}
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.