下一个具有相同数字的最大数字

Question

my code seems right, but i need it to return -1 if no bigger number can be generated:

def next_bigger(n):
    strNum = str(n)
    length = len(strNum)
    for i in range(length-2, -1, -1):
        current = strNum[i]
        right = strNum[i+1]
        if current < right:
            temp = sorted(strNum[i:])
            next = temp[temp.index(current) + 1]
            temp.remove(next)
            temp = ''.join(temp)    
            return int(strNum[:i] + next + temp)
        else: 
            return -1
    return n

My attempt to solve this isn't working: adding the else is what I percevied to be the alternative to when current is greater than right .

Please help!

Answer 1

Anyway, the flow of your code is wrong: in your loop, you have the following structure:

for A :
    if B :
        return
    else :
        return

So your program will always terminate before a second iteration.

Answer 2

In some cases fixing the code is much harder than rewritting it. I am not sure of how much help this is to you but try the following code.

def next_bigger(n):
    str_num = str(n)
    size = len(str_num)
    for i in range(2, size + 1):
        sublist = list(str_num[-i:size])
        temp = sorted(sublist, reverse=True)
        if sublist != temp:
            return int(str_num[:size-i] + ''.join(temp))
    return -1

What it does is that it slices the number from the back (starting with 2 element slices and going on up to len ) and checks to see if the generated slice produces the biggest number possible when joined . If not it gets replaced with the next bigger and returns it. Let me know if this worked for you.

---

EXAMPLE

n = 4181536841

sublist = ['4', '1']
temp = ['4', '1']  # they are the same so no larger number can be produced just by looking at a slice of length 2.

#---------iteration 2---------------
sublist = ['8', '4', '1']
temp = ['8', '4', '1']  # they are again the same so no larger number can be produced just by looking at a slice of length 3.

#---------iteration 3---------------
sublist = ['6', '8', '4', '1']
temp = ['8', '6', '4', '1']  # now they are different. So produce a number out of temp (8641) ans stich it to the rest of the number (418153)
return 4181538641

Answer 3

Ev. Kounis is wrong. In eg next biggest after 4181536841 is 4181538146, not 4181538641.

Logic works like this:

1 - Find where list[-n] > list[-n - 1]
2 - find next number higher than list[-n] after list[-n]
3 - Switch next highest and list[-n - 1]
4 - reorder the remaining numbers after list[-n] from low to high

e.g. 29357632:
1 - list[-n] = 7 --> 2935[7]632
    since 7 > 5

2 - next highest number after list[-n] (7) is the 6 --> 29357[6]32

3 - switch list[-n-1] and next highest number (switch 5 and 6) --> 
293[6]7[5]32

4 - reorder rest of digits after list[-n-1] (6) --> 2936[7532] > 29362357

so next highest number after 29357632 is 29362357