[英]Ascending and Descending Number Order in java
我在java中執行升序和降序編號,這是我的代碼:
System.out.print("Enter How Many Inputs: ");
int num1 = Integer.parseInt(in.readLine());
int arr[] = new int[num1];
for (int i = 0; i<num1; i++) {
System.out.print("Enter Value #" + (i + 1) + ":");
arr[i] =Integer.parseInt(in.readLine());
}
System.out.print("Numbers in Ascending Order:" );
for(int i = 0; i < arr.length; i++) {
Arrays.sort(arr);
System.out.print( " " +arr[i]);
}
System.out.println(" ");
System.out.print("Numbers in Descending Order: " );
目前,代碼生成以下內容:
Enter How Many Inputs: 5
Enter Value #1:3
Enter Value #2:5
Enter Value #3:6
Enter Value #4:11
Enter Value #5:2
Numbers in Ascending Order: 2 3 5 6 11
Numbers in Descending Order:
因此, Arrays.sort(arr)
調用似乎有效 - 但我正在尋找一種類似的簡單方法來提供降序排序,並且無法在文檔中找到它。 有任何想法嗎?
我想到了三種可能的解決方案:
1.撤銷訂單:
//convert the arr to list first
Collections.reverse(listWithNumbers);
System.out.print("Numbers in Descending Order: " + listWithNumbers);
2.向后迭代並打印出來:
Arrays.sort(arr);
System.out.print("Numbers in Descending Order: " );
for(int i = arr.length - 1; i >= 0; i--){
System.out.print( " " +arr[i]);
}
3.用“oposite”比較器對其進行排序:
Arrays.sort(arr, new Comparator<Integer>(){
int compare(Integer i1, Integer i2) {
return i2 - i1;
}
});
// or Collections.reverseOrder(), could be used instead
System.out.print("Numbers in Descending Order: " );
for(int i = 0; i < arr.length; i++){
System.out.print( " " +arr[i]);
}
public static void main(String[] args) {
Scanner input =new Scanner(System.in);
System.out.print("enter how many:");
int num =input.nextInt();
int[] arr= new int [num];
for(int b=0;b<arr.length;b++){
System.out.print("enter no." + (b+1) +"=");
arr[b]=input.nextInt();
}
for (int i=0; i<arr.length;i++) {
for (int k=i;k<arr.length;k++) {
if(arr[i] > arr[k]) {
int temp=arr[k];
arr[k]=arr[i];
arr[i]=temp;
}
}
}
System.out.println("******************\n output\t accending order");
for (int i : arr){
System.out.println(i);
}
}
}
你為什么要使用array
並打擾想要數字的第一個問題?
首選與相應比較器關聯的ArrayList
:
List numbers = new Arraylist();
//add read numbers (int (with autoboxing if jdk>=5) or Integer directly) into it
//Initialize the associated comparator reversing order. (since Integer implements Comparable)
Comparator comparator = Collections.reverseOrder();
//Sort the list
Collections.sort(numbers,comparator);
你可以為升序創建兩個函數,另一個用於將數組轉換為List后降序下兩個函數
public List<Integer> sortDescending(List<Integer> arr){
Comparator<Integer> c = Collections.reverseOrder();
Collections.sort(arr,c);
return arr;
}
下一個功能
public List<Integer> sortAscending(List<Integer> arr){
Collections.sort(arr);
return arr;
}
package pack2;
import java.util.Scanner;
public class group {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner data= new Scanner(System.in);
int value[]= new int[5];
int temp=0,i=0,j=0;
System.out.println("Enter 5 element of array");
for(i=0;i<5;i++)
value[i]=data.nextInt();
for(i=0;i<5;i++)
{
for(j=i;j<5;j++)
{
if(value[i]>value[j])
{
temp=value[i];
value[i]=value[j];
value[j]=temp;
}
}
}
System.out.println("Increasing Order:");
for(i=0;i<5;i++)
System.out.println(""+value[i]);
}
int arr[] = { 12, 13, 54, 16, 25, 8, 78 };
for (int i = 0; i < arr.length; i++) {
Arrays.sort(arr);
System.out.println(arr[i]);
}
像以前一樣對數組進行排序,但是使用一個倒計時而不是向上計數的循環以相反的順序打印元素。
此外,將循環移出循環 - 當前只需要對其進行一次排序時,您正在反復對數組進行排序。
Arrays.sort(arr);
for(int i = 0; i < arr.length; i++){
//Arrays.sort(arr); // not here
System.out.print( " " +arr[i]);
}
for(int i = arr.length-1; i >= 0; i--){
//Arrays.sort(arr); // not here
System.out.print( " " +arr[i]);
}
只需按升序對數組進行排序並向后打印即可。
Arrays.sort(arr);
for(int i = arr.length-1; i >= 0 ; i--) {
//print arr[i]
}
您可以先對數組進行排序,然后在兩個方向上循環兩次:
Arrays.sort(arr);
System.out.print("Numbers in Ascending Order:" );
for(int i = 0; i < arr.length; i++){
System.out.print( " " + arr[i]);
}
System.out.print("Numbers in Descending Order: " );
for(int i = arr.length - 1; i >= 0; i--){
System.out.print( " " + arr[i]);
}
Arrays.sort(arr, Collections.reverseOrder());
for(int i = 0; i < arr.length; i++){
System.out.print( " " +arr[i]);
}
並將Arrays.sort()
移出for循環..你在每次迭代時對相同的數組進行排序..
您可以采用相反的順序獲取升序數組和輸出,因此將第二個for語句替換為:
for(int i = arr.length - 1; i >= 0; i--) {
...
}
如果你在類路徑上有Apache的commons-lang,它有一個你可以使用的方法ArrayUtils.reverse(int [])。
順便說一句,您可能不希望在for循環的每個循環中對其進行排序。
使用reverse for循環以降序打印,
for (int i = ar.length - 1; i >= 0; i--) {
Arrays.sort(ar);
System.out.println(ar[i]);
}
我是以這種方式完成的(我是java的新手(也是編程))
import java.util.Scanner;
public class SortingNumbers {
public static void main(String[] args) {
Scanner scan1=new Scanner(System.in);
System.out.print("How many numbers you want to sort: ");
int a=scan1.nextInt();
int i,j,k=0; // i and j is used in various loops.
int num[]=new int[a];
int great[]= new int[a]; //This array elements will be used to store "the number of being greater."
Scanner scan2=new Scanner(System.in);
System.out.println("Enter the numbers: ");
for(i=0;i<a;i++)
num[i] = scan2.nextInt();
for (i=0;i<a;i++) {
for(j=0;j<a;j++) {
if(num[i]>num[j]) //first time when executes this line, i=0 and j=0 and then i=0;j=1 and so on. each time it finishes second for loop the value of num[i] changes.
k++;}
great[i]=k++; //At the end of each for loop (second one) k++ contains the total of how many times a number is greater than the others.
k=0;} // And then, again k is forced to 0, so that it can collect (the total of how many times a number is greater) for another number.
System.out.print("Ascending Order: ");
for(i=0;i<a;i++)
for(j=0;j<a;j++)
if(great[j]==i) System.out.print(num[j]+","); //there is a fixed value for each great[j] that is, from 0 upto number of elements(input numbers).
System.out.print("Discending Order: ");
for(i=0;i<=a;i++)
for(j=0;j<a;j++)
if(great[j]==a-i) System.out.print(+num[j]+",");
}
}
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