QUICKSORT : how to resolve java.lang.StackOverflowError?

Question

I am writing a quicksort program to run on an input size of 100000. I have tried running it at a size of 500 and it works fine but with million inputs, the program breaks with the following error code

"java.lang.StackOverflowError"

Can someone please help me how to resolve this problem? I am pretty sure that I am not caught in an infinite recursion. There is a base case which should cause the recursive method to return.

public class count_comparisons {

    public static int count_comp =0;
    public static int partitioning(int[] A, int lo, int hi) {

        int pivot = A[lo];

        int i=lo+1;
        int j=lo+1;
        int k=lo;

        for ( j=lo+1;j<=hi;j++) {            
            if (A[j] < pivot) {
                swap(A,i,j);
                i++;
            }
        }
        swap(A,i-1,lo);        
        return i-1;
    }

    public static int quicksort(int[] A, int lo, int hi) {
        if (lo>=hi) return 0;
        int pivot = partitioning(A,lo,hi);

        //StdOut.println("Pivot index is "+ pivot +" and entry at pivot is " + A[pivot]);

        StdOut.println("Lo is "+ lo +" and Hi is " + hi);
        int h = quicksort(A,lo,pivot-1);
        int m = quicksort(A,pivot+1,hi);
        //StdOut.println("First half count is "+h);
        //StdOut.println("Second half count is "+m);
        count_comp = count_comp + h + m;
        return (hi-lo);
    }

    public static void quicksort(int[] A,int N) {  
        int k = quicksort(A,0,N-1);
        count_comp = count_comp + k;
        //StdOut.println(" First count is "+k);
    }

    private static void swap(int[] A, int j,int k) {
        int temp = A[j];
        A[j] = A[k];
        A[k] = temp;
    }

    public static void main(String[] args) {
        In in = new In("input_file.txt"); 
        int N=569;
        int[] A = new int[569];
        int i=0;
        while (!in.isEmpty()) {
            A[i++] = in.readInt();
        }
        count_comparisons.quicksort(A,N);

        for( int h=0;h<N;h++) {}
            //StdOut.print(A[h]);
        StdOut.println();
        StdOut.println(count_comparisons.count_comp);

    }
}

Answer 1

Recursion does not need to be infinite in order to cause stack overflow: all it needs it to be just long enough to overflow the stack.

Quicksort may be very slow: under some particularly unfortunate circumstances, it may take up to n-1 invocations, for the worst-case performance of O(n^2) .

You have two choices - rewrite the code without recursion by using an explicit stack data structure, or by increasing the size of stack that JVM allocates to your program's threads.

Answer 2

尾部递归消除和仅递归到较小的子集是有技巧的，这限制了递归深度。