394. 字符串解码

class Solution {
public:
  string decodeString(string s) {
    string Result;
    string TokenCount;
    stack<int> Counts;
    stack<string> SubStrings;

    for (char c : s) {
      if (isdigit(c))
        TokenCount += c;
      if (isalpha(c)) {
        // We need to discriminate top-level expressions.
        if (Counts.empty())
          Result += c;
        else
          SubStrings.top() += c;
      }
      // Push one level.
      if (c == '[') {
        Counts.push(stoi(TokenCount));
        SubStrings.emplace();
        // Clear TokenCount immediately. Otherwise nested expression would fail.
        TokenCount.clear();
      }
      // Pop one level.
      if (c == ']') {
        int Repeat = Counts.top();
        string Processed;

        for (int i = 0; i < Repeat; ++i)
          Processed += SubStrings.top();
        Counts.pop();
        SubStrings.pop();
        // Put Processed string onto previous level. Still, we need to check if
        // we're at the top level.
        if (Counts.empty())
          Result += Processed;
        else
          SubStrings.top() += Processed;
      }
    }
    return Result;
  }
};