Java中的递归程序 [英] Recursive program in Java

问题描述

我有一个程序可以实现这一点:

I have a program that realizes this:

Input: "(-123)+(200)" -> String
Output: 77 -> Int

Input: "((-123)+(200))*((2)+(1))"
Output: 231

好吧，我有代码，但我不知道使用正则表达式是否是个好主意.老师给了我一个分隔String的函数，例如:

Well, I have the code, but I don't know if it's a good idea to use regular expressions. The teachers gave me a function that separates the String, for example:

int n = StringUtils.exprMainOperator ("(-123)+(200)");
n = 7;

我的想法是将每个字符串用 'n' 分隔在一个子字符串中，但我不知道如何递归地实现这一点.任何想法?请不要在代码中，所以我可以学习，谢谢.

My idea is to separate, by 'n' each String, in a substring, but I don't know how to realize this recursively. Any idea? Not in code please, so I can learn, thanks.

推荐答案

实现它的方法有很多种，这是一些可以工作的伪代码.我假设您只有算术运算符，并且 StringUtils.exprMainOperator 处理括号/操作优先级:

There are many ways of implementing it, this is some pseudocode that would work. I am assuming that you only have arithmetic operators, and that StringUtils.exprMainOperator handles parenthesis/priority of operations:

public static int evaluateExpression(String exp) {
    if(exp.charAt(0) == '(' && exp.charAt(exp.length()-1) == ')' 
        && /*These two parenthesis correspond to each other*/) return evaluateExpression(exp.substring(1, exp.length()-1));

    if(/*is Valid Number*/) return Integer.parseInt(exp);

    int n = StringUtils.exprMainOperator(exp);
    char op = exp.charAt(n);
    String preop = exp.substring(0, n), postop = exp.substring(n+1);

    if(op == '+') return evaluateExpression(preop) + evaluateExpression(postop);
    if(op == '-') return evaluateExpression(preop) - evaluateExpression(postop);
    if(op == '*') return evaluateExpression(preop) * evaluateExpression(postop);
    if(op == '/') return evaluateExpression(preop) / evaluateExpression(postop);

    //You shouldn't reach this part of the code
    return -1;
}

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