Python-使用唯一数字查找最近的更大数字

Question

I came across this problem in which you will take an integer as a input from the user and then return the nearest greater number with unique digits .
First it seem like an easy one and i wrote it's code which gave me an desired outputs but for some inputs returns an integer with repeating digits or a unique number with higher value then expected,
I want to know why my code is showing different behaviour than expected and what would be the right answer to this problem. Also i don't know what to do when a digit will become two digit number how to make it unique eg. 9999

• code

    n = int(input("enter a no.:"))+1         #taking input adding 1 to make sure if the input is 
                                         #unique it should return higher no. with uniq 
    a =[]                                    #creating an empty list
    while n!=0:                              #loop to store digits in a list
        a.append(n%10)                       #at index 0 ones digit is stored
        n = int(n/10)                        #at index 1 tens digit is stored an so on...


    while len(set(a)) != len(a):             #checking if the list is unique
        for i in range(0,len(a)):            #
            occur = a.count(a[i])            #taking occurence of a list item
            if occur != 1:                   #if no. is repeated
                a[i]+=occur-1                #incrementing the repeating digit.

    a.reverse()                              #converting list to integer and printing
    n = 0
    for i in a:
        n = n*10+i
    print(n)

• Behaviour
  1.Printing repeating digits for some inputs
  2.Printing Higher values than expected for some inputs
  3.When a digit becomes a two digit no. it treats it like a single digit
• Some outputs

enter a no.:1233  
output:  1234            #desired output: 1234

enter a no.:7885
output:  7896            #desired output: 7890

enter a no.:7886
output:  8008             #desired output: 7890

enter a no.:999
output:  2013             #desired output: 1023

Answer 1

You're probably over complicating this for yourself.

Would something like this work rather than doing all of the conversions?

n = int(input("enter a no.:"))+1   #taking input adding 1 to make 
                                   #sure if the input is unique then 
                                   #program does't return the input itself

a = str(n)                         # simply convert to a string

while len(set(a)) != len(a):       #checking if the string is unique
    n += 1
    a = str(n)

print(n)

Answer 2

Why not simply increment the number until you find one with unique digits?

def next_uniq(n):
  a = str(n)  
  while len(set(a)) != len(a):
    a = str(int(a) + 1)
  return a

for i in [1233, 7885, 7886, 999]:
  print(next_uniq(i))

# 1234, 7890, 7890, 1023