递归算法查找数组中的最小元素

Question

I have some pseudo code for a recursive algorithm that finds the smallest number in an array.

Here is the algorithm.

Min(A[0..n - 1])
If n = 1 return A[0]
else
{ 
  temp <-- Min(A[0..n - 2])
  if temp <= A[n - 1]
     return temp
  else return A[n - 1]
}

One part I don't understand about this pseudo code is the line "temp <-- Min(A[0..n - 2])". Specifically why is it "n - 2" in the recursive call instead of "n - 1"?

My other question is how I would implement that line in code. I am using Java.

Thanks in advance for any help.

Answer 1

Because you array is indexed from 0 to n-1, inclusive. You need to recurse on a sub-array that's one element smaller, hence n-2.

Answer 2

for your 1st question:

since you are using a recursive algorithm, in order to solve the problem, you first have to solve the same problem with a smaller size. In this pseudocode, for finding the minimum of an array with the length of n, first the minimum of the same array with the size of n-1 is found and then the minimum is compared with nth element. Your array is indexed from 0 to n-1 (which will make it's length = n) so for recursive call, you have to call the array from index 0 to n-2 (n-1 elements).

for your 2nd question: This is how I would implement the code in Java:

public class Minimum {
   public Minimum(int[] A) {
    findMin(A, 0, A.length-1);
}

public int findMin(int [] A, int start, int end){
    if (start== end-1)
        return A[0];
    else{
        int temp=findMin(A, start, end-1 );
        if (temp<=A[end]) 
            return  temp;
        else 
            return A[end];
    }
}

}

Answer 3

My other question is how I would implement that line in code.

If you are using an array.

// create an array with one less element.
A a2 = new A[a.length-1];
// copy the elements
System.arrayCopy(a,0,a2,0,a2.length);

If you are using a List

List<A> a2 = a.subList(0, a.length-1);