如何使用递归在使用 python 的列表中给出最小 integer？

Question

I am trying to write a recursive function called my_minimum that receives a list of integers as parameter and returns the minimum stored in the list. I am able to get the user to input integers separated by a space. However, it gives me a number totally unrelated to what I am trying to achieve.

For example if user list was: 67 89 45 34 23 3 45 67 78 , it should give me give me 3 instead of 23.

def my_minimum(A, n):
    # if size = 0 means whole list
    # also if it has been traversed
    if (n == 1):
        return A[0]
    return min(A[n - 1], my_minimum(A, n - 1))

# Driver Code
if __name__ == '__main__':
    input_string= input("Enter a list element separated by space: ")
    A  = input_string.split()
    n = len(A)
    print(my_minimum(A, n))

Answer 1

This is a string comparison issue, not an issue with your recursion. The string "3" compares as greater than the string "23" , just like "B" sorts after "AA" in lexicographical order.

Try converting your strings to integers if you want integer ordering:

A  = list(map(int, input_string.split()))