python 3.2 - find second smallest number in a list using recursion

Question

So I need to find the second smallest number within a list of integers using recursion but I cannot for the life of me devise a way to do it. I can do it with to find smalle

Answer 1

Here is a pure recursive way. You need to return the first smallest and the second smallest element in a tuple and so on.

def smallest(int_list):

    if(len(int_list) == 2):
        if (int_list[0] >= int_list[1]):
            return (int_list[0],int_list[1])
        else:
            return (int_list[1],int_list[0])
    else:
        first_smallest,second_smallest = smallest(int_list[1:])
        current_elem = int_list[0]
        if(second_smallest <= current_elem):
            return (second_smallest,current_elem)
        else:
            if (current_elem<=first_smallest):
               return (current_elem,first_smallest)
            else:
               return (first_smallest,second_smallest)



if __name__ == "__main__":    
    small = smallest([1,2,3,4])
    print small[1]
    //output 2