你可以把你的问题分解成3个部分：
1.首先创建两个列表，一个列表按遇到的顺序包含偶数，另一个列表包含奇数：

private static List<List<Integer>> createOddityLists(int... numbers) {
        List<Integer> numsList = Arrays.stream(numbers).boxed().collect(Collectors.toList());
        
        List<List<Integer>> numsByOddity = new ArrayList<List<Integer>>();
        numsByOddity.add(new ArrayList<>()); // List of odd numbers
        numsByOddity.add(new ArrayList<>()); // List of even numbers
        
        numsList.forEach(num -> numsByOddity.get(num % 2).add(num));
        return numsByOddity;
    }

1.用零（0 s）填充两个列表中较短的一个，使其长度与另一个相等：

private static void padShorterList(List<List<Integer>> numsByOddity) {
        int sizeDiff = numsByOddity.get(0).size() - numsByOddity.get(1).size();
        int listIndexToBePadded = sizeDiff < 0 ? 0 : 1;
        List<Integer> padding = Collections.nCopies(Math.abs(sizeDiff), 0);
        numsByOddity.get(listIndexToBePadded).addAll(padding);
    }

1.最后，将这两个列表交织在一起：

private static List<Integer> joinLists(List<List<Integer>> numsByOddity) {
        List<Integer> resultList = new ArrayList<>(numsByOddity.get(1));
        for (int idx = 0; idx < numsByOddity.get(0).size(); idx++)
            resultList.add(idx * 2, numsByOddity.get(0).get(idx));
        return resultList;
    }

以下是完整的工作示例：

public class ArrayRearrangement {
    
    public static void main(String[] args) {
//      int[] result = rearrange(1, 2, 3, 4);
        int[] result = rearrange(1, 1, 1, 4);
        System.out.println(Arrays.stream(result).boxed().collect(Collectors.toList()));
    }
    
    private static int[] rearrange(int... numbers) {
        List<List<Integer>> numsByOddity = createOddityLists(numbers);
        padShorterList(numsByOddity);
        return joinLists(numsByOddity).stream().mapToInt(i->i).toArray();
    }

    private static List<List<Integer>> createOddityLists(int... numbers) {
        List<Integer> numsList = Arrays.stream(numbers).boxed().collect(Collectors.toList());
        
        List<List<Integer>> numsByOddity = new ArrayList<List<Integer>>();
        numsByOddity.add(new ArrayList<>()); // List of odd numbers
        numsByOddity.add(new ArrayList<>()); // List of even numbers
        
        numsList.forEach(num -> numsByOddity.get(num % 2).add(num));
        return numsByOddity;
    }

    private static void padShorterList(List<List<Integer>> numsByOddity) {
        int sizeDiff = numsByOddity.get(0).size() - numsByOddity.get(1).size();
        int listIndexToBePadded = sizeDiff < 0 ? 0 : 1;
        List<Integer> padding = Collections.nCopies(Math.abs(sizeDiff), 0);
        numsByOddity.get(listIndexToBePadded).addAll(padding);
    }

    private static List<Integer> joinLists(List<List<Integer>> numsByOddity) {
        List<Integer> resultList = new ArrayList<>(numsByOddity.get(1));
        for (int idx = 0; idx < numsByOddity.get(0).size(); idx++)
            resultList.add(idx * 2, numsByOddity.get(0).get(idx));
        return resultList;
    }
}

Complete code on GitHub

使用数组我们可以做这样的事情，代码需要优化。

public static int[]  arrangeInEvenOddOrder(int[] arr)
    {
        // Create odd and even arrays
        int[] oddArr = new int[arr.length];
        int[] evenArr = new int[arr.length];
        int oCount = 0, eCount = 0;
        // populate arrays even and odd
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] % 2 == 0)
                evenArr[eCount++] = arr[i];
            else
                oddArr[oCount++] = arr[i];
        }
        int[] resArr = new int[oCount >= eCount?
                               2*oCount :  2*eCount-1];
        // populate elements upto min of the
        // two arrays
        for (int i =0; i < (oCount <= eCount?
                2*oCount : 2*eCount ); i++ )
        {
            if( i%2 == 0)
                resArr[i] = evenArr[i/2];
            else
                resArr[i] = oddArr[i/2];
        }
        // populate rest of elements of max array
        // and add zeroes
        if (eCount > oCount)
        {
            for (int i=2*oCount,j=0;i<2*eCount-1; i++)
            {
                if (i%2 == 0)
                {
                    resArr[i] = evenArr[oCount+j];
                    j++;
                }
                else
                    resArr[i] = 0;
            }
        }
        else if (eCount < oCount)
        {
            for (int i=2*eCount,j=0;i<2*oCount; i++)
            {
                if ( i%2 != 0)
                {
                    resArr[i] = oddArr[eCount+j];
                    j++;
                }
                else
                    resArr[i] = 0;
            }
        }
        return resArr;
    }

根据索引对元素排序，即如果元素是偶数，则必须位于偶数位置，反之亦然

int sortArrayByEvenOddIndex(int arr[]) {
     int n = arr.length;
     int res[] = new int[n];
     int odd = 1;
     int even = 0;
    
     for (int i = 0; i < n; i++) {
         if (arr[i] % 2 == 0) {
             res[even] = arr[i];
             even += 2;
         } else {
             res[odd] = arr[i];
             odd += 2;
         }
     }
     return  res;
 }