I've been wanting to practice my python and so I want to be able to locate the number 3 using recursion but the code itself continues to run past the number of elements in the list. I had used a while loop but that hadn't work and neither does a generic if statement
listofnum = [7,6,21,12,3,99,8,3,0]
def recursion(x,counter):
if counter >= 0:
if x[0] == 3:
print("here")
recursion(x[1:],len(x)-1)
else:
print("next item...")
recursion(x[1:],len(x)-1)
else:
return "Done"
It does find 3 both times, but the code is in an endless loop
I would suggest instead of passing a new list every time( by slicing the list), pass same list and update the counter
value.
listofnum = [7,6,21,12,3,99,8,3,0]
def recursion(x, current_index=0):
if current_index >= len(x): # base condition
return
if x[current_index] == 3:
print("Got 3 at:", current_index)
recursion(x, current_index+1)
You need to return the counter when you find 3, otherwise, slice the list and continue.
listofnum = [7,6,21,12,3,99,8,3,0]
def recursion(x,counter):
if counter > 0:
if x[0] == 3:
return counter
else:
print("next item...")
return recursion(x[1:],len(x)-1)
else:
return -1
res = recursion(listofnum, len(listofnum))
print(res)
It's not clear how you invoke your recursion function, but I guess it's in form like:
recursion(listofnum, len(listofnum))
In my case, trying your code, it correctly stops with an
IndexError: list index out of range error
because of the " counter >= 0
" condition, which makes your code attempting to read on position 0
even if the supplied list has no elements at all, thus no 0 (first) position does exist.
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