Java: Find the Kth largest element of an array, using 2 arrays

问题

Here is the description about the algorithm/solution:

Input: An array of ints and an array index
Output: Kth largest element of the array
Read the first K elements into an auxiliary array (the K largest found so far)
Sort the K-element array

Then:

 for each remaining element {
    if (if it is smaller than the smallest element of the aux. array) {
        throw it away
    } else { 
        remove the current smallest element of the auxiliary array
        place the element into the correct position in the auxiliary array
    }
 }

 then return the smallest (the Kth) element of the auxiliary array

I have come up with the following solution:

public int findElement() throws IndexingError {
    int[] bigArray = getArray();
    int k = getIndex();
    if (k <= 0 || k > bigArray.length) {
        throw new IndexingError();
    }

    int[] smallArray = Arrays.copyOfRange(bigArray,0,k-1);
    Arrays.sort(smallArray);


    for (int i = k; i < bigArray.length; i++) {
        for (int ii = 1; ii < smallArray.length; ii++) {
            smallArray[ii-1] = smallArray[ii];
            if (bigArray[i] > smallArray[ii]) {
                smallArray[ii] = bigArray[i];
                System.out.println(smallArray[ii] + " " + smallArray[ii-1] + " " + bigArray[i]);
                break;

            }
        }

    }

    return smallArray[0];
}

So for example:

The code should return the value of: 8, based on the info below:

array = [2, 3, 8, 7, 1, 6, 5, 9, 4, 0]
k = 2

Running the above code yields different outputs every time, if you run it enough times you get the correct answer?

Could someone identify/fix flaws in my code?

回答1:

You need to fix the following issues:

In Arrays.copyOfRange(array, from, to) method the to index is exclusive. So, you need to have it like Arrays.copyOfRange(bigArray,0,k) in order to copy first k elements to the auxiliary array.
The logic where you loop over the remaining array elements from the main array and update the auxiliary array is wrong.

For example, let auxiliary array be: [4, 6], and the next main array element be 2. According to your logic the auxiliary array will be updated to [6, 6] since you first copy the element at index 1 to index 0 in the auxiliary array and then check if the new element(from main array) is greater than the element at index 1 in the auxiliary array. In this case the new element is smaller, so just the copying takes place, and we get corrupted auxiliary array.

What you simply need to do is, for every new element from the main array, check if it's greater than the first element in the auxiliary array. If it is, then this new element should be a part of the auxiliary array and should be placed at the proper position. You can use the bubble sort technique to find the proper place.

public int findElement() throws IndexingError {
    int[] bigArray = getArray();
    int k = getIndex();
    if (k <= 0 || k > bigArray.length) {
        throw new IndexingError();
    }

    int[] smallArray = Arrays.copyOfRange(bigArray,0,k);
    Arrays.sort(smallArray);


    for (int i = k; i < bigArray.length; i++) {
        if(bigArray[i] > smallArray[0]) {
            smallArray[0] = bigArray[i];
            int j = 0;
            while((j < k-1) && (smallArray[j] > smallArray[j+1])) {
                swap(smallArray[j], smallArray[j+1]);
                j++;
            }
        }

    }

    return smallArray[0];
}

来源：https://stackoverflow.com/questions/52690295/java-find-the-kth-largest-element-of-an-array-using-2-arrays

标签

java

arrays

sorting

for-loop

int