def rev(n):
'''Return the result of reversing the digits in integer n. For example,
rev(512) should return 215.'''
passdef rev(n):
'''Return the result of reversing the digits in integer n. For example,
rev(512) should return 215.'''
n = str(n)[-1] + str(n)[:-1]def rev(n):
reversed = str(n)[-1] + str(n)[:-1]
return reversed
n = 5112
print rev(n)dacruick said:Can you give me all of the bounds of your assignment? There are most likely python objects that can do this problem and save you from having to use recursion yourself, but I assume you're not allowed to do that.
Mark44 said:I'm not very knowledgeable about python, but I think this would work.
I'm assuming that the parameter n is an int.
1) Calculate digit = n % 10. This gives you the current digit in the 1's place.
2) Print digit.
3) Calculate a new value for n: n = n/10. Note that this is integer division, so for example 12/10 == 1 and 8/10 == 0.
4) If n > 0, call rev(n). If n == 0, end the recursion.
Actually, the logic is very pure.dacruick said:Thats a cool way to do it. This way you actually don't have to use any other variable other than n. I didn't even think about that. However, I don't think this is conceptually the easiest code to write. It's based more on nuances of programming rather than pure logic.
Mark44 said:I'm not very knowledgeable about python, but I think this would work.
I'm assuming that the parameter n is an int.
1) Calculate digit = n % 10. This gives you the current digit in the 1's place.
2) Print digit.
3) Calculate a new value for n: n = n/10. Note that this is integer division, so for example 12/10 == 1 and 8/10 == 0.
4) If n > 0, call rev(n). If n == 0, end the recursion.
zeion said:So I can only print the last digit each time it recurses?
How can I return the result as one whole string?
dacruick said:You could simply create another variable and add the digit on each time.
You could also create an indexed list and concatenate it.
So instead of the print digit, it would be digit = digit + newdigit.
the only limitation there is that it would have to be a string.
def rev(n):
'''Return the result of reversing the digits in integer n. For example,
rev(512) should return 215.'''
n = str(n)
if not n:
return (n)
else:
return rev(n[1:]) + n[0]Recursion in Python is a programming technique where a function calls itself repeatedly until a certain condition is met. This allows for solving complex problems by breaking them down into smaller, simpler problems.
In Python, a recursive function calls itself with a smaller input until it reaches the base case, which is the simplest form of the problem. Once the base case is reached, the function stops calling itself and returns a result. The results from each recursive call are then combined to solve the original problem.
Recursion and iteration are both ways to achieve repetitive execution in Python. The main difference is that recursion involves a function calling itself, while iteration uses loops to repeat a block of code. Recursion can be more elegant and concise, but it can also be less efficient and use more memory compared to iteration.
Recursion can help solve complex problems in a simpler, more elegant way by breaking them down into smaller, more manageable problems. It can also lead to more concise and readable code. Additionally, some problems are inherently recursive in nature, making recursion the most efficient and logical approach to solving them.
One common mistake is forgetting the base case, which can lead to an infinite loop. Another mistake is not properly breaking down the problem into smaller subproblems, resulting in inefficient code. Additionally, too many recursive calls can lead to stack overflow, which can be avoided by optimizing the recursive function.