Question

我知道在SO上已经有类似的问题，但我找不到解决我遇到的问题的问题。我正在尝试创建一个方法，将中缀符号表达式转换为后缀表示法，同时实现运算符的优先级以获得正确的输出。我用通常的方法（push，pop，peek等）制作了我自己的堆栈类，它的工作非常好。我的问题是，对于更复杂的表达式，如A-（B + C ^ D ^ C）/ D * B，我得到错误的输出。转换的结果应该是ABCDC ^^ + D / B * - 而我继续获得ABCDC ^^ + D / -B

这是我的方法：

    public static String infixToPostfix(char[] expressionArray, CharStack opStack){
    String output = "";
    int length = expressionArray.length;
    for(int i = 0; i < length; i++){    
        if(isOperatorOrBracket(expressionArray[i])){ 
            if(priorityAtInput(expressionArray[i]) >= priorityAtStack(opStack.peek())){
                opStack.push(expressionArray[i]);
            }else if(priorityAtInput(expressionArray[i]) == priorityAtStack(opStack.peek())){
                output = output + expressionArray[i];
            }else{
                while(opStack.peek() != '('){
                    output = output + opStack.pop();
                }
                opStack.pop();
            }
        }else{
            output = output + expressionArray[i];
        }
    }
    while(!opStack.empty()){
        if(opStack.peek() != '('){
            output = output + opStack.pop();
        }else if(opStack.peek() == '('){
            opStack.pop();
        }
    }
    return output;
}

如果您需要任何组件方法，请告诉我。任何帮助非常感谢！

Answer 1

经过一个小时的盯着屏幕后，我发现了问题。非常感谢eclipse中的调试器！

public static String infixToPostfix(char[] expressionArray, CharStack opStack){
    String output = "";
    int length = expressionArray.length;
    for(int i = 0; i < length; i++){    
        if(isOperatorOrBracket(expressionArray[i])){ 
            if(priorityAtInput(expressionArray[i]) >= priorityAtStack(opStack.peek())){
                opStack.push(expressionArray[i]);
            }else if(priorityAtInput(expressionArray[i]) < priorityAtStack(opStack.peek())){
                while(priorityAtInput(expressionArray[i]) < priorityAtStack(opStack.peek())){
                    output = output + opStack.pop();
                    if(opStack.peek() == '('){
                        opStack.pop();
                        break;
                    }else if(priorityAtInput(expressionArray[i]) >= priorityAtStack(opStack.peek())){
                        opStack.push(expressionArray[i]);
                        break;
                    }
                }
            }else{
                while(opStack.peek() != '('){
                    output = output + opStack.pop();
                }
                opStack.pop();
            }
        }else{
            output = output + expressionArray[i];
        }
    }
    while(!opStack.empty()){
        if(opStack.peek() != '('){
            output = output + opStack.pop();
        }else if(opStack.peek() == '('){
            opStack.pop();
        }
    }
    return output;
}

使用堆栈和运算符优先级的中缀到Postfix

1 个答案: