java - PostFix 评估结果错误

Question

public static int postFixEvaluation(int[] numValues, String postfix){
            Stack<Integer> evaluateStack = new Stack<Integer>();
            char[] chars = postfix.toCharArray();
            int length = chars.length;
            int currentNumValue =0; 
            int currentLocation =0; 
            for(int i = 0; i < length; i++){
                char currentChar = chars[i];
            if(Character.isLetter(currentChar))//checks to see if character is a letter 
            {
                //replace all letters with values 
                currentLocation = charToNum(currentChar);//this retrieves the location of specific letter
                currentNumValue= (numValues[currentLocation]);//retrieves the value of that location
                evaluateStack.push(currentNumValue);//get the number value of that variable and push it on stack
                System.out.println(Arrays.toString(evaluateStack.toArray()));//prints out stack elements 

            }

                else if(isOperator(currentChar)){//checks if character is an operator 
                    switch(currentChar){//switches evaluation according to operator 
                    case '+': evaluateStack.push(evaluateStack.pop() + evaluateStack.pop()); break;
                    case '*': evaluateStack.push(evaluateStack.pop() * evaluateStack.pop()); break;
                    case '-': evaluateStack.push(evaluateStack.pop() - evaluateStack.pop()); break;
                    case '/': evaluateStack.push(evaluateStack.pop() / evaluateStack.pop()); break;
                    }
                }

            }
            if (!evaluateStack.isEmpty()) //as long as the stack is not empty 
                return evaluateStack.pop();//returns the result
            else
                return 0;//if it is empty returns zero 
        }

输入:

C = -13
X = 5
H = 25
D = 4
$PART2
C-(B+(C+(A-E)))-X
(D+C)-(H*X-C)
(A-H)/(D+B)
A*H+C Infix: C-(B+(C+(A-E)))-X
Postfix: CBCAE-++-X-
[-13]
[-13, 2]
[-13, 2, -13]
[-13, 2, -13, 1]
[-13, 2, -13, 1, 5]
[6, 5]
Result: -1**Result should be: -3**

Infix: (D+C)-(H*X-C)
Postfix: DC+HX*C--
[4]
[4, -13]
[-9, 25]
[-9, 25, 5]
[-9, 125, -13]
Result: -129 **Result should be: -147**

Infix: (A-H)/(D+B)
Postfix: AH-DB+/
[1]
[1, 25]
[24, 4]
[24, 4, 2]
Result: 0**Result should be: -4**


Infix: A*H+C
Postfix: AH*C+
[1]
[1, 25]
[25, -13]
Result: 12**Correct Result**

我不明白为什么我的 postEvaluation 不会产生正确的输出。如果有人可以请帮忙!我在我的代码中进行了大量评论，所以如果我能澄清任何事情，请告诉我。谢谢!

Answer 1

我手边没有电脑，只有旧式的纸笔、手机和一些关于调车场算法的知识。

我认为你的错误是这样的(只在纸上测试过):

您不是在评估 A-E，而是在进行 E-A。其他运营商也是如此。

错误:

evaluateStack.push(evaluateStack.pop() + evaluateStack.pop());

正确:

//Don't change pop order here!
int righthand = evaluateStack.pop();
int lefthand = evaluateStack.pop();

evaluateStack.push(lefthand + righthand);

这是我更正的内容:

public static int postFixEvaluation(int[] numValues, String postfix){
    Stack<Integer> evaluateStack = new Stack<Integer>();
    char[] chars = postfix.toCharArray();
    int length = chars.length;
    int currentNumValue =0;
    int currentLocation =0;

    for (int i = 0; i < length; i++){
        char currentChar = chars[i];

        //checks to see if character is a letter
        if (Character.isLetter(currentChar)){
            //replace all letters with values 
            currentLocation = charToNum(currentChar);//this retrieves the location of specific letter
            currentNumValue = (numValues[currentLocation]);//retrieves the value of that location
            evaluateStack.push(currentNumValue);//get the number value of that variable and push it on stack
            System.out.println(Arrays.toString(evaluateStack.toArray()));//prints out stack elements 
        }


        //checks if character is an operator 
        if (isOperator(currentChar)){
            int righthand = evaluateStack.pop();
            int lefthand = evaluateStack.pop();
            switch (currentChar){
            //switches evaluation according to operator 
                case '+': evaluateStack.push(lefthand + righthand); break;
                case '*': evaluateStack.push(lefthand * righthand); break;
                case '-': evaluateStack.push(lefthand - righthand); break;
                case '/': evaluateStack.push(lefthand / righthand); break;
            }
        }
    }
    if (!evaluateStack.isEmpty()){ //as long as the stack is not empty
        return evaluateStack.pop();//returns the result
    } else {
        return 0; //if it is empty returns zero 
    }
}

一些额外的提示:您应该已经添加了您的辅助函数以及代码在算法方面对您的问题的作用。这可能会吸引更多人来回答您的问题。