java - 使用堆栈中缀到 Postfix

Question

我不确定我这样做是否正确。我正在尝试将中缀方程转换为后缀方程，例如:

(3 + 4) * 2 

在后缀中是:

4 3 + 2 *

如果可能的话，我试图用一种方法来完成这一切。

现在我收到一个 arrayoutofbounds 错误，所以我弹出错误的地方或什么的。

这是 infixtopostfix 方法。

public void InfixToPostfix(String f) {
    Stacked postfix = new Stacked(100);
    Stacked op = new Stacked(100);
    char c;
    String fix = f;
    //String out = "";
    int out = 0;

    for (int i = 0; i < fix.length(); i++) {
        c = fix.charAt(i);

        if (c != '+' || c != '*' || c != '-' || c != '/' || c != '(' || c != ')') {
            // out += c;
            postfix.push(c);
        } 
        else if (c == '+' || c == '*' || c == '-' || c == '/') {
            if (c != ')') {
                op.push(c);
            } else {
                out = (char) op.pop();
                s.push(out);
            }
            out = (char) op.pop();
            System.out.println("THE POSTFIX = " + out)
        }
    }
}

Answer 1

  public class Postfix {

   private int priority(char ch) {
    if (ch == '^')
     return 3;
    if (ch == '/' || ch == '*')
     return 2;
    if (ch == '+' || ch == '-')
     return 1;
    return 0;
   }

   public String toPostfix(String in ) {

    String copy = in +")";
    Stack s = new Stack(copy.length());
    s.push('(');

    int i, l = copy.length();
    char ch;

    String r = "";
    for (i = 0; i < l; i++) {

     ch = copy.charAt(i);
     if (Character.isLetter(ch) == true)
      r += ch;

     else if (ch == '(')
      s.push(ch);

     else if (ch == ')') {
      while (s.seeTop() != '(')
       r += s.popSeeTop();

      s.pop();

     } else {
      while (priority(ch) <= priority(s.seeTop()))
       r += s.popSeeTop();

      s.push(ch);
     }
    }
    return r;
   }
  }