Question

我不确定我是否正确行事。我正在尝试将中缀方程转换为后缀方程，如：

(3 + 4) * 2

后缀中的

是：

4 3 + 2 *

如果可能，我试图在一种方法中完成所有这些。

现在我收到了一个arrayoutofbounds错误，所以我在错误的地方或其他地方弹出。

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;
   }
  }

Answer 2

public class InfixToPostfix {

Stack operatorStack = new Stack();

public String toPrefix(String expression) {
    expression="("+expression+")";
    int i;      
    char token;
    String output = "";
    for (i = 0; i < expression.length(); i++) {
        token = expression.charAt(i);
        if (Character.isLetterOrDigit(token) == true)
            output += token;
        else if (token == '(')
            operatorStack.push(token);
        else if (token == ')') {
            char seeTop;
            while ((seeTop = seeTop()) != '(') {
                output += seeTop;
                popSeeTop();
            }

            operatorStack.pop();
        } else {
            while (priority(token) <= priority(seeTop())) {
                output += seeTop();
                popSeeTop();
            }
            operatorStack.push(token);
        }
    }
    return output;

}

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

private Character seeTop() {
    if(!operatorStack.empty())
        return (Character) operatorStack.peek();
    else
        return '0';
}

private void popSeeTop() {
    if(!operatorStack.empty())
        operatorStack.pop();
}

Answer 3

/**
 * Convert arithmetic from infix to postfix
 *
 * @example 1 * ( 12 + 23 ) - ( 4 / 5 ) => 1 12 23 + * 4 5 / -
 * @author  Yong Su
 */
import java.util.Stack; 
import java.lang.StringBuilder;

class InfixToPostfix {

    public static void main(String[] args) {
        String infix = "1 * ( 12 + 23 ) - ( 4 / 5 )";
        String postfix = convert(infix);
        System.out.println(postfix);
    }

    /**
     * Perform the conversion
     *
     * @param expression Arithmetic infix format
     */
    public static String convert(String expression) {
        // Use StringBuilder to append string is faster than
        // String concatenation
        StringBuilder sb = new StringBuilder();

        // Use a stack to track operations
        Stack<Character> op = new Stack<Character>();

        // Convert expression string to char array
        char[] chars = expression.toCharArray();

        // The length of expression character
        int N = chars.length;

        for (int i = 0; i < N; i++) {
            char ch = chars[i];

            if (Character.isDigit(ch)) {
                // Number, simply append to the result
                while (Character.isDigit(chars[i])) {
                    sb.append(chars[i++]);
                }
                // Use space as delimiter
                sb.append(' ');
            } else if (ch == '(') {
                // Left parenthesis, push to the stack
                op.push(ch);
            } else if (ch == ')') {
                // Right parenthesis, pop and append to the result until meet the left parenthesis
                while (op.peek() != '(') {
                    sb.append(op.pop()).append(' ');
                }
                // Don't forget to pop the left parenthesis out
                op.pop();
            } else if (isOperator(ch)) {
                // Operator, pop out all higher priority operators first and then push it to the stack
                while (!op.isEmpty() && priority(op.peek()) >= priority(ch)) {
                    sb.append(op.pop()).append(' ');
                }
                op.push(ch);
            }
        }

        // Finally pop out whatever left in the stack and append to the result
        while(!op.isEmpty()) {
            sb.append(op.pop()).append(' ');
        }

        return sb.toString();
    }

    /**
     * Check if a character is an operator
     */
    private static boolean isOperator(char ch) {
        return ch == '^' || ch == '*' || ch == '/' || ch == '+' || ch == '-';
    }

    /**
     * Define the operator priority
     */
    private static int priority(char operator) {
        switch (operator) {
            case '^' : return 3;
            case '*' :
            case '/' : return 2;
            case '+' :
            case '-' : return 1;
        }
        return 0;
    }

}

Answer 4

使用简单的Stacks

，这是您尝试做的一个特别优雅和简单的实现

public class InfixToPostfix {
    public static void main(String[] args) {
        Stack<String> stack = new Stack<String>();
        while (!StdIn.isEmpty()) {
            String s = StdIn.readString();
            if      (s.equals("+")) stack.push(s);
            else if (s.equals("*")) stack.push(s);
            else if (s.equals(")")) StdOut.print(stack.pop() + " ");
            else if (s.equals("(")) StdOut.print("");
            else                    StdOut.print(s + " ");
        }
        StdOut.println();
    }
}

使用堆栈的中缀到Postfix

4 个答案: