下面寫的代碼是正確的，但我想縮短這個代碼。用java編寫一個程序，在單維數組中輸入10個數字，并以這樣的方式排列它們，即所有偶數后面都跟著所有奇數。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]);}我可以將偶數和奇數排列在單個 for 循環中，而不是兩個 for 循環中嗎？我使用兩個for循環將偶數和奇數轉換為數組。請縮短代碼。一個用于循環遍歷，用于檢查偶數，第二個用于奇數。b[]
4 回答
慕尼黑5688855
TA貢獻1848條經驗 獲得超2個贊
這是一個簡單的程序給你。
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);
}
}
慕婉清6462132
TA貢獻1804條經驗 獲得超2個贊
在這個版本中，它將偶數復制到開頭，將奇數復制到結尾。
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));
}
指紋
[2, 4, 6, 10, 9, 7, 3, 1]
回首憶惘然
TA貢獻1847條經驗 獲得超11個贊
此代碼將幫助您分離偶數和奇數。
// 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);
}
}
撒科打諢
TA貢獻1934條經驗 獲得超2個贊
我建議閱讀流，它們將使收集處理變得更加容易
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));
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