结合QuickSort和中值选择算法

Question

I want to modify QuickSort (in Java) so that every time Partition is called, the median of the proportioned array is used as the pivot.

I have a median selection algorithm in Java that returns the kth smallest element, in this case the median. I have tons of quicksort algorithms in java that all work by themselves and sort an array. Unfortunately I can't combine those two in order to achieve the above... Everytime I try it i usually get stackoverflow erros.

Can anybody show me code to see how it can be done?

Thanks

EDIT: For example this is a median selection algorithm that I have tried to use.

public int quickSelect(int[] A, int p, int r, int k) {
    if (p==r) return A[p];
    int q = Partition(A,p,r);
    int len = q-p+1;

    if (k == len) return A[q];
    else if (k<len) return Select(A,p,q-1,k);
    else return Select(A,q+1,r,k-len);
}

public int partition(int[]A, int p, int r) {
    int x = A[r];
    int i = p-1;
    for (int j = p; j<=r-1; j++) {
        if (A[j] <= x) {
            i++;
            swap(A,i,j);
        }
    }
    swap(A,i+1,r);
    return i+1;
}

It works by itself but when I try to call quickSelect through quicksort's partition function to return the pivot to be used, it doesn't work. Obviously I'm doing something wrong but I don't know what. Unfortunately on the Internet I haven't found any algorithm, even in pseudocode, that would combine a median selection with quicksort.

Answer 1

The standard way to get the median is to sort the data. And you want to sort the data by partitioning on the median. This seems very chicken and egg to me.

Could you elaborate on why you want to partition/pivot on the median?

Answer 2

What you are looking for is the Selection Algorithm. Here's a link with pseudocode .

From the link:

In computer science, a selection algorithm is an algorithm for finding the kth smallest number in a list

To find the median you want to find the k=floor((n+1)/2) smallest number in the list where n is the size of the list.

Answer 3

Note that in PARTITION the pivot is A[r] .

public int QUICKSORT2(int[] A, int p, int r) {
    if (p<r) {
        int median=Math.floor((p + r) /2) - p + 1
        int q=SELECT(A, p, r, median)
        q=PARTITION2(A, p, r, q)
        QUICKSORT2(A, p, q-1)
        QUICKSORT2(A, q+1, r)
    }
}

public int PARTITION2(int[]A, int p, int r, int q) {
    int temp = A[r];
    A[r]=A[q];
    A[q]=temp;
    return PARTITION(A, p, r)
}

Answer 4

You can use this ...

int Select(int array[],int start, int end,int k){

if(start==end){
    return start;
}

int x=array[end];
int i=start-1;
for(int j=start;j<=end-1;j++){
    if(array[j]<x){
        i++;
        Swap(array+i,array+j);
    }
}
i++;
Swap(array+i,array+end);

if(i==k){
    return i;
}
else if(i>k){
    return Select(array,start,i-1,k);
}
else{
    return Select(array,i+1,end,k);
}

}

Select will partition array on kth smallest element in array;