Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in

the integer n and is greater in value than n. If no such positive 32-bit integer exists, you need to return -1.

思路：

求元素的全排列，然后选出比n大的所有元素中最小的那个。

vector
     
      digit;    int backup = n;    while(n)    {        digit.push_back(n%10);        n/=10;    }    sort(digit.begin(),digit.end());    long long result = INT_MAX +1LL;        do {                long long temp =0;                for(int i=0;i
      
        backup) result = min(result, temp);        } while (next_permutation(digit.begin(), digit.end()));    if(result<=INT_MAX) return result;    else return -1;