用调车场算法评估RPN的Java语言

我在网上找到了调车场算法：

import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

public class ShuntingYardAlgorithm {

private enum Operator {
    ADD(1), SUBTRACT(2), MULTIPLY(3), DIVIDE(4);
    final int precedence;
    Operator(int p) {
        precedence = p;
    }
}

private Map<String, Operator> operatorMap = new HashMap<String, Operator>() {/**
     * 
     */
    private static final long serialVersionUID = 1L;

{
    put("+", Operator.ADD);
    put("-", Operator.SUBTRACT);
    put("*", Operator.MULTIPLY);
    put("/", Operator.DIVIDE);
}};

private boolean isHigherPrec(String op, String sub) {
    return (operatorMap.containsKey(sub) &&
            operatorMap.get(sub).precedence >= operatorMap.get(op).precedence);
}

public String shuntingYard(String infix) {
    StringBuilder output = new StringBuilder();
    Stack<String> stack = new Stack<String>();

    for (String token : infix.split("")) {
        //operator
        if (operatorMap.containsKey(token)) {
            while ( ! stack.isEmpty() && isHigherPrec(token, stack.peek())) {
                output.append(stack.pop()).append(' ');
            }
            stack.push(token);
        }
        //left parenthesis
        else if (token.equals("(")) {
            stack.push(token);
        }
        //right parenthesis
        else if (token.equals(")")) {
            while ( ! stack.peek().equals("(")) {
                output.append(stack.pop()).append(' ');
            }
            stack.pop();
        }
        //digit
        else {
            output.append(token).append(' ');
        }
    }

    while ( ! stack.isEmpty()) {
        output.append(stack.pop()).append(' ');
    }

    return output.toString();
}

例如我输入：

256+3

结果：

2 5 6 3+

---

评估：

6+3=9，忽略2和5

当遇到操作员或括号时，您的

调车场

函数将丢弃

输出

的内容

在处理当前字符之前，需要添加对输出内容的检查

    if (operatorMap.containsKey(token)) {
        // TODO: Check output here first, and create a token as necessary
        while ( ! stack.isEmpty() && isHigherPrec(token, stack.peek())) {
            output.append(stack.pop()).append(' ');
        }
        stack.push(token);
    }
    //left parenthesis
    else if (token.equals("(")) {
        // TODO: Check output here first, and create a token as necessary
        stack.push(token);
    }
    //right parenthesis
    else if (token.equals(")")) {
        // TODO: Check output here first, and create a token as necessary
        while ( ! stack.peek().equals("(")) {
            output.append(stack.pop()).append(' ');
        }
        stack.pop();
    }

此外，使用空字符串进行拆分相当于一次只迭代字符串一个字符。使用

toCharArray（）

迭代

infix

可能更具可读性

   for (char c : infix.toCharArray())

---

你为什么给我看这个？我问的是分隔符，不是算法。

    if (operatorMap.containsKey(token)) {
        // TODO: Check output here first, and create a token as necessary
        while ( ! stack.isEmpty() && isHigherPrec(token, stack.peek())) {
            output.append(stack.pop()).append(' ');
        }
        stack.push(token);
    }
    //left parenthesis
    else if (token.equals("(")) {
        // TODO: Check output here first, and create a token as necessary
        stack.push(token);
    }
    //right parenthesis
    else if (token.equals(")")) {
        // TODO: Check output here first, and create a token as necessary
        while ( ! stack.peek().equals("(")) {
            output.append(stack.pop()).append(' ');
        }
        stack.pop();
    }

   for (char c : infix.toCharArray())