Largest Number After Digit Swaps by Parity LeetCode solution in C++

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class Solution {
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 public:
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  int largestInteger(int num) {
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    priority_queue<int> p;        // priority queue to store odd digits in descending order
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    priority_queue<int> q;        // priority queue to store even digits in descending order
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    string nums = to_string(num); // converting num to a string for easy access of the digits
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    int n = nums.size();          // n stores the number of digits in num
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    for (int i = 0; i < n; i++) {
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      int digit = nums[i] - '0';
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      if ((digit) % 2) // if digit is odd, push it into priority queue p
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        p.push(digit);
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      else
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        q.push(digit); // if digit is even, push it into priority queue q
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    }
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    int answer = 0;
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    for (int i = 0; i < n; i++) {
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      answer = answer * 10;
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      if ((nums[i] - '0') % 2) // if the digit is odd, add the largest odd digit
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                               // of p into the answer
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      {
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        answer += p.top();
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        p.pop();
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      } else {
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        answer += q.top();
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        q.pop();
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      } // if the digit is even, add the largest even digit of q into the answer
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    }
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    return answer;
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  }
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};