CODEWITHSUNDEEP
pythonpython-3.xintegeriterationreverse
Anil
Anil

Reputation: 13

write a python program to reverse the given numbers

Input=12345 output=54321

Program to write a reverse of numbers

n=int(input("Enter number :"))
rev=0
while(n>0):
    dig = n%10
    rev=rev*10+dig
    n=n//10
print("The reverse of the number:" ,rev)

Can some explain why this print 54321

Upvotes: 0

Views: 5320

Answers (6)

adsa
adsa

Reputation: 53

You can iterate a string in python, and you'll be iterating by each character

There you go:

a = input("Enter : ")
x = [num for num in a]  # Getting the individual digits
b = reversed(x)         # reversed() is a built in function
print("".join(b))

Upvotes: 2

python
python

Reputation: 36

This is much simpler

inp = input("Enter the number")
list1 = [numbers for numbers in inp]
list2 = [reversed_numbers for reversed_numbers in list1[::-1]]
print("".join(list2))

Upvotes: 1

Timur Shtatland
Timur Shtatland

Reputation: 12337

See other answer for great explanations. Here I add 2 simpler solutions to the original problem - how to reverse a number:

# For example, 12345:
num = input("Enter number: ")

# fastest and shortest:
rev = num[::-1]

# or (slower and longer)
rev = "".join(reversed(num))

print(f"The reverse of the number: {rev}")
# The reverse of the number: 54321

Upvotes: 0

Petr Synek
Petr Synek

Reputation: 323

It's actualy pretty simple

1st iteration

  1. Well you first get % of 10 which is basicaly last number (residual) dig = 5,
  2. than you add this to rev*10 (in first iteration 0) + dig = 5,
  3. and you whole nuber divide the n by 10 (deleting last nuber) n = 1234

2nd iteration

  1. dig = 4
  2. rev =5*10 + 4 = 54
  3. n = 123

3rd iteration

  1. dig = 3
  2. rev = 540 + 3 = 543
  3. n = 12

...

However there is much easier way how to do this eg.

If your input is string so you just need to reverse it

input = "12345"
output = input[::-1]

If input is number (int, float) than you have to convert it to string

input = 12345
output = str(input)[::-1]

And convert back

int(output)

Upvotes: 0

Suvajit Patra
Suvajit Patra

Reputation: 190

Let, n = 12345 , where 12345 is the given input.

rev = 0
Now in the first iteration of the while loop n > 0
dig = remainder of n divided by 10, So it is = 5
rev = rev(0) * 10 + dig(5) = 5
n = n // 10 = 1234

In the second iteration n = 1234 which is > 0

rev = 5
dig = remainder of n divided by 10, So it is = 4
rev = rev(5) * 10 + dig(4) = 54
n = n // 10 = 123

In the third iteration n = 123 which is > 0

rev = 54
dig = remainder of n divided by 10, So it is = 3
rev = rev(54) * 10 + dig(3) = 543
n = n // 10 = 12

In the fourth iteration n = 12 which is > 0

rev = 543
dig = remainder of n divided by 10, So it is = 2
rev = rev(543) * 10 + dig(2) = 5432
n = n // 10 = 1

In the fifth iteration n = 1 which is > 0

rev = 5432
dig = remainder of n divided by 10, So it is = 1
rev = rev(5432) * 10 + dig(1) = 54321
n = n // 10 = 0

Now in the sixth iteration of the while loop n = 0 so it is not greater than 0 so loop will break and you got rev = 54321

Upvotes: 0

CopyrightC
CopyrightC

Reputation: 863

This should work

number = int(input("Enter a number: "))
rev = ''
 
while number > 0:
    last_digit = number % 10 #The remainder when it's divided by 10 will always be the last digit
    rev = rev + str(last_digit)
    number = number // 10
 
print("Reversed : ", rev)

Upvotes: 0

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