交换数组中的奇数和偶数

Question

我在这个网站上看到了这段代码。它使用一种方法对数组进行排序，偶数排在前面，奇数排在数组后面。我想知道你是否可以做同样的事情，只是让奇数先出现，然后偶数出现？我试过了但无济于事。我是 java 编码的新手，我想测试递归。

public class Recurse {

//i=left 
//j=right
//first i tried to shift the whole thing
//have all even numbers pop to the front of array when even
public static int[] seperator(int[] arr1, int left, int right){
    int temp;
    if(left>=right)       //base case, return array
        return arr1; 
    else if(arr1[left]%2!=0 && arr1[right]%2==0){//if match, do the swap
        temp=arr1[left];
        arr1[left]=arr1[right];
        arr1[right]=temp;   
        return seperator(arr1, left+1, right-1);
    }
    else{
        if(arr1[right]%2!=0){//if right side is on odd #, then decrease index
            return seperator(arr1, left, right-1);
        }
        if(arr1[left]%2==0){//if left side is on even #, then increase index
            return seperator(arr1, left+1, right);
        }
    } 
    return arr1;
}

public static void main(String[] args){

    //int index=0;
    int[] arry={70,13,48,19,24,5,7,10};

    int[] newarry=seperator(arry, 0, arry.length-1);
    System.out.print("The new sorted array is: ");
    for(int i=0; i<newarry.length;i++){
        System.out.print(newarry[i]+" ");
    } 
}

}

输出是这样的：

  The new sorted array is: 70 10 48 24 19 5 7 13

Answer 1

好吧，如果您希望奇数排在第一位，请将任何 %2!=0 替换为 %2==0 并将任何 %2==0 替换为 %2!=0 :

  public static int[] seperator(int[] arr1, int left, int right){
    int temp;
    if(left>=right)       
      return arr1; 
    else if(arr1[left]%2==0 && arr1[right]%2!=0){
      temp=arr1[left];
      arr1[left]=arr1[right];
      arr1[right]=temp;   
      return seperator(arr1, left+1, right-1);
    }
    else{
      if(arr1[right]%2==0){//if right side is on even #, then decrease index
        return seperator(arr1, left, right-1);
      }
      if(arr1[left]%2!=0){//if left side is on odd #, then increase index
        return seperator(arr1, left+1, right);
      }
    } 
    return arr1;
  }

Answer 2

试试这样的东西：

public static int[] seperator(int[] arr1, int left, int right) {
    int temp;
    if (left >= right) // base case, return array
        return arr1;
    else if (arr1[left] % 2 == 0 && arr1[right] % 2 != 0) {// if match, do the swap
        temp = arr1[left];
        arr1[left] = arr1[right];
        arr1[right] = temp;
        return seperator(arr1, left + 1, right - 1);
    } else {
        if (arr1[right] % 2 == 0) {// if right side is on odd #, then decrease index
            return seperator(arr1, left, right - 1);
        }
        if (arr1[left] % 2 != 0) {// if left side is on even #, then increase index
            return seperator(arr1, left + 1, right);
        }
    }
    return arr1;
}
Output:
The new sorted array is: 7 13 5 19 24 48 70 10

Answer 3

public class Program {

    public static void main(String[] args) {
        int a[] = { 2, 3, 5, 6, 7, 8, 10, 46, 50 };

        for (int i = 0; i < a.length - 1; i++) {
            for (int j = 0 ; j < a.length-1; j++) {
                if (a[j] % 2 != 0) {
                    int temp = a[j];
                    a[j] = a[j + 1];
                    a[j + 1] = temp;
                }
            }
        }

        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
    }

}

交换数组中的奇数和偶数

Swap odd and even numbers in array

java

recursion