Infix to Postfix Expression

最新推荐文章于 2021-11-26 16:45:34 发布

weixin_30737363

最新推荐文章于 2021-11-26 16:45:34 发布

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CC 4.0 BY-SA版权

文章标签： python

原文链接：http://www.cnblogs.com/beiyeqingteng/p/11225354.html

本文介绍了一种将中缀表达式转换为后缀表达式的算法，并提供了详细的步骤说明及Java实现代码。该算法通过扫描输入的中缀表达式，利用栈来辅助处理运算符的优先级，最终输出等价的后缀表达式。

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Example :

Infix : (A+B) * (C-D) )

Postfix: AB+CD-*

算法：

1. Scan the infix expression from left to right.

2. If the scanned character is an operand, append it to result.

3. Else

　　3.1 If the precedence of the scanned operator is greater than the precedence of the operator in the stack(or the stack is empty or the stack contains a ‘(‘ ), push it.

　　3.2 Else, Pop all the operators from the stack which are greater than or equal to in precedence than that of the scanned operator. After doing that Push the scanned operator to the stack. (If you encounter parenthesis while popping then stop there and push the scanned operator in the stack.)

4. If the scanned character is an ‘(‘, push it to the stack.

5. If the scanned character is an ‘)’, pop the stack and and output it until a ‘(‘ is encountered, and discard both the parenthesis.

6. Repeat steps 2-6 until infix expression is scanned.

7. Append the remaing items int the stack.

public class InfoxToPostfix {
    private int order(char ch) {
        switch (ch) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        case '^':
            return 3;
        default:
            return -1;
        }
    }

    private String infixToPostfix(String exp) {
        StringBuilder result = new StringBuilder();
        Stack<Character> stack = new Stack<>();

        for (int i = 0; i < exp.length(); ++i) {
            char c = exp.charAt(i);
            if (Character.isLetterOrDigit(c)) {
                result.append(c);
            } else if (c == '(') {
                stack.push(c);
            } else if (c == ')') {
                while (!stack.isEmpty() && stack.peek() != '(') {
                    result.append(stack.pop());
                }
                stack.pop();
            } else {
                while (!stack.isEmpty() && order(c) <= order(stack.peek())) {
                    result.append(stack.pop());
                }
                stack.push(c);
            }
        }
        while (!stack.isEmpty()) {
            result.append(stack.pop());
        }
        return result.toString();
    }
}