Question

下面写的代码是正确的，但是我想缩短这段代码。

用Java编写程序，以在一维数组中输入10个数字，并以所有偶数后跟所有奇数的方式排列它们。

int a[] = new int[6];
int b[] = new int[6];
int i, j;
int k = 0;
System.out.println("enter array");
for (i = 0; i < 6; i++) {  
    a[i] = sc.nextInt();
}
for (j = 0; j < 6; j++) {
    if (a[j] % 2 == 0) {
        b[k] = a[j];
        k++;
    }
}
for (j = 0; j < 6; j++) {
    if (a[j] % 2 != 0) {
        b[k] = a[j];
        k++;
    }
}
System.out.println("out-put");
for (i = 0; i < 6; i++) {  
    System.out.println(b[i]);
}

我可以在一个for循环中而不是两个for循环中排列偶数和奇数吗？我正在使用两个for循环将偶数和奇数传输到b[]数组中。请缩短代码。一个用于循环遍历以检查偶数，另一个用于检查奇数。

Answer 1

这是一个简单的程序。

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;

/**
 *
 * @author Momir Sarac
 */
public class GroupByEvenAndOddNumbers {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // create a collection
        List<Integer> listOfNumbers = new ArrayList<>();
        // do code within a loop for 10 times
        for(int i=0;i<10;i++)
        {
            //print to screen this text
            System.out.println("Input your number:");
            //get next input integer
            int number = scanner.nextInt();
            // add it to collection
            listOfNumbers.add(number);
        }
        // sort this collection, list of numbers
        // convert all numbers(positive and negative ) within to 0 or 1 depending whether or not they are even or odd and sort them accordignaly.
        Collections.sort(listOfNumbers, Comparator.comparingInt(n -> Math.floorMod(n, 2)));
        //print sorted collection
        System.out.println("Ordered list ..." + listOfNumbers);
    }
}

Answer 2

在此版本中，它将偶数复制到开头，将奇数复制到结尾。

static int[] sortEvenOdd(int... nums) {
    int even = 0, odd = nums.length, ret[] = new int[nums.length];
    for (int num : nums)
        if (num % 2 == 0)
            ret[even++] = num;
        else
            ret[--odd] = num;
    return ret;
}

public static void main(String[] args) {
    int[] arr = {1, 3, 2, 4, 7, 6, 9, 10};
    int[] sorted = sortEvenOdd(arr);
    System.out.println(Arrays.toString(sorted));
}

打印

[2, 4, 6, 10, 9, 7, 3, 1]

Answer 3

此代码将帮助您分隔偶数和奇数。

// java code to segregate even odd 
// numbers in an array 
public class GFG { 

// Function to segregate even 
// odd numbers 
static void arrayEvenAndOdd( 
            int arr[], int n) 
{ 

    int i = -1, j = 0; 
    while (j != n) { 
        if (arr[j] % 2 == 0) 
        { 
            i++; 

            // Swapping even and 
            // odd numbers 
            int temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        j++; 
    } 

    // Printing segregated array 
    for (int k = 0; k < n; k++) 
        System.out.print(arr[k] + " "); 
} 

// Driver code 
public static void main(String args[]) 
{ 
    int arr[] = { 1, 3, 2, 4, 7, 
                        6, 9, 10 }; 
    int n = arr.length; 
    arrayEvenAndOdd(arr, n); 
 } 
}

Answer 4

由于您没有必要将偶数和奇数本身分别在数组的一半进行排序，因此您可以在输入时将它们分配给关联的数组部分。因此，您只需要使用两个“计数器”变量，一个在左边，从零开始递增，一个在右边，从数组长度减一开始递减，一个递增。然后，您可以添加您的数字，检查是否为偶数，将左计数器增加，则分配为奇数；如果奇数，右计数器减少，则分配。循环执行此操作，直到您的左计数器大于右计数器。我创建了一个简单的示例，在将String解析为int时，我没有检查NumberFormatException：

import java.util.Arrays;
import java.util.Scanner;

public class SortedArrayInput {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter length of array: ");
        final int arrayLength = Integer.parseInt(scanner.nextLine());

        int intArray[] = new int[arrayLength];
        for (int l = 0, r = arrayLength - 1; l <= r; ) {
            System.out.print("Enter new array value: ");
            int v = Integer.parseInt(scanner.nextLine());
            intArray[v % 2 == 0 ? l++ : r--] = v;
        }

        System.out.println("Output: " + Arrays.toString(intArray));
    }
}

示例输入/输出：

Enter length of array: 6
Enter new array value: 1
Enter new array value: 2
Enter new array value: 3
Enter new array value: 4
Enter new array value: 5
Enter new array value: 6
Output: [2, 4, 6, 5, 3, 1]

Answer 5

我建议阅读流，它们将使您的收集处理变得容易得多

List<Integer> numbers = new ArrayList<>();
        numbers.add(1);
        numbers.add(2);
        numbers.add(3);
        numbers.add(4);
        numbers.add(5);
        numbers.add(6);
        numbers.add(7);
        numbers.add(8);
        numbers.add(9);
        numbers.add(0);

        //this way you simply traverse the numbers twice and output the needed ones
        System.out.println(numbers.stream()
                .filter(x->x%2==0)
                .collect(Collectors.toList()));
        System.out.println(numbers.stream()
                .filter(x->x%2==1)
                .collect(Collectors.toList()));

        //this way you can have the numbers in two collections
        numbers.forEach(x-> x%2==0? addItToEvenCollection : addItToOddCollection);

        //this way you will have a map at the end. The boolean will tell you if the numbers are odd or even, 
        // and the list contains the numbers, in order of apparition in the initial list
        numbers.stream().collect(Collectors.groupingBy(x->x%2==0));

Answer 6

一种检查数字是否为偶数的高性能方法是使用如果（（x＆1）== 0）

显示偶数后跟所有奇数

6 个答案: