leetcode：556. Next larger element III [simulation + change as little as possible]

analysis

If it has reached the maximum, return -1
otherwise , There must be a number in front that is smaller than that in the back , We exchange it
Then we need to make the front number as far back as possible
After the exchange , We need to rearrange the numbers after this number （ Ensure the minimum ascending order ）
Also pay attention to 32 Bit integer range , otherwise -1

ac code

class Solution:
    def nextGreaterElement(self, n: int) -> int:
        flag = False
        s = str(n)
        if len(s) == 1:
            return -1
        for i in range(1, len(s)):
            if s[i] > s[i - 1]:
                flag = True

        if not flag:
            return -1
        
        lst = list(s)
        idx = 0

        for i in range(len(lst)):
            for j in range(i):
                if lst[len(s) - 1 - j] > lst[len(s) - 1 - i]:
                    
                    lst[len(s) - 1 - j], lst[len(s) - 1 - i] = lst[len(s) - 1 - i], lst[len(s) - 1 - j]
                    flag = False
                    idx = len(s) - 1 - i
                    break
            if not flag:
                break
        
        # idx Rearranging behind 
        temp = lst[idx + 1:]

        temp.sort()
        lst[idx + 1:] = temp[:]

        ans = int(''.join(lst))
        return ans if ans <= 2 ** 31 - 1 else -1

summary

Next spread
Find the two minimum numbers that make the interaction larger （ The first one should try to stay back , The latter one can also be as far back as possible ）
Then you can remake the first one in ascending order