1

我正在使用 Java 开发一个项目，该项目需要我将中缀表达式转换为后缀表达式。我目前可以使用这种方法将中缀表达式转换为后缀，只要它们不包含括号，但我不知道如何处理括号。

基本上，我有两个堆栈，其中包含称为“令牌”的对象。Token 是一个包装类，它包含一个字符串，该字符串可以是数字、变量（被评估为数字，等待用户输入）、运算符（运算符具有与其关联的优先级，以便我的方法可以确定如何处理“+”、“-”、“*”和“/”之间的操作顺序）或括号（括号有办法确定它是开括号还是闭括号）。

我应该如何处理括号？多层括号呢？

public String toPostFix() {
    StringBuilder postfixstr = new StringBuilder();

    Stack<Token> in_fix = new Stack<>();
    Stack<Token> post_fix = new Stack<>();

    for (int i = tokens.length - 1; i >= 0; i--) {
        t = new Token(tokens[i]);
        in_fix.push(t);
    }

    //there are still tokens to process
    while (!in_fix.empty()) {
        //is a number
        if (in_fix.peek().type == 1) {     
            postfixstr.append(in_fix.pop().toString());
        } 

        //is an operator and the stack is empty
        else if (in_fix.peek().type == 3 && post_fix.empty()) {   
            post_fix.push(in_fix.pop());
        } 

        // is an operator that has higher priority than the operator on the stack
        else if (in_fix.peek().type == 3 && in_fix.peek().isOperator() > post_fix.peek().isOperator()) {
            post_fix.push(in_fix.pop());
        } 

        // is an operator that has lower priority than the operator on the stack
        else if (in_fix.peek().type == 3 && in_fix.peek().isOperator() <= post_fix.peek().isOperator()) {
            postfixstr.append(post_fix.pop());
            post_fix.push(in_fix.pop());
        } 

        //puts the rest of the stack onto the output string
        if (in_fix.empty()) {
            while (!post_fix.empty()) {
                postfixstr.append(post_fix.pop());
            }
        }
    }

    return postfixstr.toString();
}
4

2 回答 2

2

您需要将左括号压入堆栈，并在遇到右括号时像这样处理堆栈：

// opening (
if (in_fix.peek().type == 4) {   
    post_fix.push(in_fix.pop());
}
//closing )
if(in_fix.peek().type == 5){
    while(!(post_fix.isEmpty() || post_fix.peek().type == 4)){
         postfixstr.append(post_fix.pop());
    }
    if (post_fix.isEmpty())
        ; // ERROR - unmatched )
    else
        post_fix.pop(); // pop the (
    in_fix.pop(); // pop the )
}
于 2013-11-01T05:47:38.380 回答
1

试试这个方法： 

    //opening Parenthesis 
        if (in_fix.peek().type == 4) {   
                    post_fix.push(in_fix.pop());
        }
        //closing Parenthesis 
        if(in_fix.peek().type == 5){
             //Till opening parenthesis encountered in stack, append operators to postfix. and pop parenthesis and do not append to post_fix.
             while(post_fix.peek().type!=4){
                 postfixstr.append(post_fix.pop());
             }
            //finally pop left parenthesis from post_fix stack.
            post_fix.pop();
        }
于 2013-11-01T05:21:20.647 回答