偶数排序，按出现递归Java的顺序先排除赔率

Question

我如何将数组排序为偶数和奇数之间的数字出现顺序相同，例如 [2 1 4 6 3 9 8] 使用递归来提供此 [1 3 9 2 4 6 8]？非常感谢一些使用 Java 的帮助或示例。

这段代码做了类似的工作，只是很难解决手头的问题。

Sorting an array recursively in Java with even numbers appearing in front of array.

这是关于如何处理并排的偶数问题的部分。不得使用循环。

public static int[] SplitOdds(int input[], int left, int next){     
    int[] temp = new int[1];
    int[] temp2 = new int[1];
    if (input.length == 1)
        {return input;}
    if (input.length ==2){
        temp[0] = input[left];
        input[left] = input[next];
        input[next] = temp[0];
        return input;}
    if (input[left]%2!=0 && input[next]%2==0)
        {return SplitOdds(input, left+1, next+1);}
    if(input[left]%2==0 && input[next]%2!=0){
        temp[0] = input[left];
        input[left] = input[next];
        input[next] = temp[0];
        return  SplitOdds(input, left+1, next+1);}
    if(input[left]%2!=0&&input[next]%2!=0)
        return SplitOdds(input, left+1, next+1);
    if(input[left]%2==0&&input[next]%2==0){
        int j = next;
        temp[0]=input[left];
        return SplitOdds(input, left, next+1);}
        else if(input[next]%2!=0){
            //temp2[0]=input[j];
            //input[left]=input[next];
            input[left+1] = temp2[0];       
        return SplitOdds(input, left, next+1);}
    else if(next == input.length)
    {left = 1; 
    next = 2;}
    return input;
    }

Answer 1

我会尝试这样的：

Integer[] array = new Integer[]{2, 1, 4, 6, 3, 9, 8}; // what to sort?

// The sorting logic is the same for anything that is Comparable. Only the criteria (ie. what makes a value bigger than another?) changes
Arrays.sort(array, new Comparator<Integer>() {
    public int compare(Integer i1, Integer i2) {
        if(i1%2 != i2%2) { // comparing an odd number with an even?
            return i1%2==0? 1:-1; // then odd comes before even
        }

        return i1 - i2; // else biggest number comes after smaller ones
    }
});

System.out.println(Arrays.toString(array)); // basic display

Answer 2

这是一种可能的方法：

  @Test
  public void test() {
    List<Integer> testSource = Arrays.asList(1, 8, 2, 4, 6, 9, 5, 3);

    Collections.sort(testSource, new Comparator<Integer>() {
      @Override
      public int compare(Integer o1, Integer o2) {
        if (o1 % 2 == o2 % 2) {
            return o1.compareTo(o2);
        } else {
          return o1 % 2 == 1 ? -1 : 1;
        }
      }
    });
    assertEquals(Arrays.asList(1, 3, 5, 9, 2, 4, 6, 8), testSource);
  }