交换数组中的奇数和偶数

Question

我在此站点上看到了此代码。它使用一种方法来对数组进行排序，其中偶数排在数组的前面，而奇数排在数组的后面。我想知道您是否可以做同样的事情，除了先显示奇数，然后再显示偶数？我试过了，但无济于事。我是Java编程的新手，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，如果你想单号是第一位的，取代任何%2!=0与%2==0任何%2==0带%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;
  }