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我稍微修改了这个算法来处理大括号和方括号

queue<string> postEvaluation(vector<string> expression){
stack<string> operators;
queue<string> output;

for(int i = 0; i < expression.size(); i++){
    if(isdigit(expression[i]))
        output.push(expression[i]);
    if(expression[i] == "");
    if (IsOperator(expression[i])) {
  /* Next, move all operators off the stack until this operator has the
   * highest precedence of them all.
   */
  while (!operators.empty() && IsOperator(operators.top()) &&
         Precedence(operators.top()) >= Precedence(expression[i])) {
    output.push(operators.top()); operators.pop();
  }

  /* Add the operator to the operator stack. */
  operators.push(expression[i]);

}

/* Case 2: Found an open parenthesis. */
    else if (expression[i] == "(" || expression[i] == "[" || expression[i] == "{" ) {
  operators.push(expression[i]);
}
/* Case 3: Found a close parenthesis. */
    else if (expression[i] == ")" || expression[i] == "]" || expression[i] == "}") {

  /* Keep moving tokens over from the operator stack to the result until
   * we find a matching open parenthesis.  If we run out of tokens, then 
   * there must be a syntax error.
   */
  while (!operators.empty() && operators.top() != "(" && !operators.empty() && operators.top() != "[" && !operators.empty() && operators.top() != "{") {
    output.push(operators.top()); operators.pop();
  }

  /* Remove the parenthesis we matched. */
  operators.pop();
}

  }

  /* Keep shifting all the operators back off the operator stack onto the end
   * of the input.  If we find an open parenthesis, then the input is
   * malformed.
   */
  while (!operators.empty()) {
    output.push(operators.top()); operators.pop();
  }
   return output;
}

出于某种原因，当我输入以下内容时： 10 - [ ( 3 - 2 ) * 10 ] / 5
输出为： 10 3 2 - 10 * - 5 /

当输出应该是：10 3 2 - 10 * 5 / -

关于如何正确修改它以处理大括号{}和括号[]的任何想法？

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0 回答 0