Given a number find the next permutation of it

This problem was asked by IBM.

Given an integer, find the next permutation of it in absolute order. For example, given 48975, the next permutation would be 49578.

My Solution(C++):

#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>


int nextHigherPermutation(int num){
  // returns the next higher permuation from the digits of a number
  std::vector<int> digits;
  while (num>0){
    digits.push_back(num%10);
    num = num/10;
  }
  int num_digits = digits.size(), i;
  // for (int i: digits) std::cout<<i<<',';std::cout<<'\n';
  std::vector<int> trail;
  for (i=0; i<num_digits; i++){
    trail.push_back(digits[i]);
    if (digits[i+1]<digits[i]) break;
  }
  int to_replace_pos = i+1;
  int to_replace = digits[to_replace_pos];
  for (int i=0; i<trail.size(); i++){
    if (trail[i]>to_replace){
      digits[to_replace_pos] = trail[i];
      trail[i] = to_replace;
      break;
    }
  }

  // std::cout<<"TRAIL\n";
  // for (int i: trail) std::cout<<i<<',';std::cout<<'\n';

  std::reverse(trail.begin(), trail.end());
  for (i = to_replace_pos; i<num_digits; i++) trail.push_back(digits[i]);

  // std::cout<<"TRAIL\n";
  // for (int i: trail) std::cout<<i<<',';std::cout<<'\n';

  int n = 0;
  for (int i=0; i<num_digits; i++){
    n = n+ trail[i]*pow(10, i);
  }
  return n;
}



void test(){
  std::cout<<nextHigherPermutation(48975)<<'\n'; //49578 [5, 7, 9, 8, 4]
  std::cout<<nextHigherPermutation(48070)<<'\n'; //48700 [0, 7, 0, 8, 4]
  std::cout<<nextHigherPermutation(121698765)<<'\n'; //121756689 [5, 6, 7, 8, 9, 6, 1, 2, 1]
}


int main(){
  test();
  return 0;
}