RPN 表达式中元素之间的空格 java

Question

我有一个方法 getRPNString()，它 returns 逆波兰表示法字符串。我想用空格键拆分这个字符串来计算它。现在我不明白如何在我的 RNP 字符串中正确添加空格键，因为它不适用于两位数字。

public class Calc1 {

public static void main(String[] args) {

    String in = "((5+3*(4+2)*12)+3)/(1+3)+5";
    String out = getRPNString(in);
    System.out.println(out);

}

private static String getRPNString(String in) {
    LinkedList<Character> oplist = new LinkedList<>();
    StringBuilder out = new StringBuilder();

    for (int i = 0; i < in.length(); i++) {
        char op = in.charAt(i);
        if (op == ')') {
            while (oplist.getLast() != '(') {
                out.append(oplist.removeLast());
            }
            oplist.removeLast();
        }

        if (Character.isDigit(op)) {

            out.append(op);

            /*int j = i + 1;
            for (; j < in.length(); j++) {
                if (!Character.isDigit(j)) {
                    break;
                }
                i++;
            }
            out.append(in.substring(i, j));*/

        }

        if (op == '(') {
            oplist.add(op);
        }

        if (isOperator(op)) {
            if (oplist.isEmpty()) {
                oplist.add(op);
            } else {
                int priority = getPriority(op);
                if (priority > getPriority(oplist.getLast())) {
                    oplist.add(op);
                } else {
                    while (!oplist.isEmpty()
                            && priority <= getPriority(oplist.getLast())) {
                        out.append(oplist.removeLast());
                    }
                    oplist.add(op);
                }
            }
        }

    }

    while (!oplist.isEmpty()) {
        out.append(oplist.removeLast());
    }

    return out.toString();
}

private static boolean isOperator(char c) {
    return c == '+' || c == '-' || c == '*' || c == '/' || c == '%';
}

private static int getPriority(char op) {
    switch (op) {

    case '*':
    case '/':
        return 3;

    case '+':
    case '-':
        return 2;

    case '(':
        return 1;

    default:
        return -1;
    }
}

}

我试图通过 append(' ') 在我的 StringBuilder 变量中添加空格键。但是两位数就不对了。我想我完全不明白怎么做。

例如，如果输入是字符串 in = "((5+3*(4+2)*12)+3)/(1+3)+5";当我向所有调用 [= 添加空格键时，输出将是 5342+12+3+13+/5+ 28=](' ')**out 是 **5 3 4 2 + * 1 2 * + 3 + 1 3 + / 5 +，所以像“12”这样的数字变成了“1 2”。你能帮忙吗？

Answer 1

只需将 Character.isDigit(op) 之后注释掉的代码更改为：

int j = i + 1;
int oldI = i;//this is so you save the old value
for (; j < in.length(); j++) {
    if (!Character.isDigit(in.charAt(j))) {
        break;
    }
    i++;
}
out.append(in.substring(oldI, j));
out.append(' ');

Answer 2

我改变了方法，现在可以正常使用了。我在写作时深知自己的错误 !Character.isDigit(j) 但需要 !Character.isDigit(in.charAt(j)).

private static String getRPNString(String in) {
    LinkedList<Character> oplist = new LinkedList<>();
    StringBuilder out = new StringBuilder();

    for (int i = 0; i < in.length(); i++) {
        char op = in.charAt(i);
        if (op == ')') {
            while (oplist.getLast() != '(') {
                out.append(oplist.removeLast()).append(' ');
            }
            oplist.removeLast();
        }

        if (Character.isDigit(op)) {

            int j = i + 1;
            int oldI = i;//this is so you save the old value
            for (; j < in.length(); j++) {
                if (!Character.isDigit(in.charAt(j))) {
                    break;
                }

                i++;
            }

            out.append(in.substring(oldI, j));
            out.append(' ');

        }

        if (op == '(') {
            oplist.add(op);
        }

        if (isOperator(op)) {
            if (oplist.isEmpty()) {
                oplist.add(op);
            } else {
                int priority = getPriority(op);
                if (priority > getPriority(oplist.getLast())) {
                    oplist.add(op);
                } else {
                    while (!oplist.isEmpty()
                            && priority <= getPriority(oplist.getLast())) {
                        out.append(oplist.removeLast()).append(' ');
                    }
                    oplist.add(op);
                }
            }
        }

    }

    while (!oplist.isEmpty()) {
        out.append(oplist.removeLast()).append(' ');
    }

    return out.toString();
}

现在可以生成正确的表达式了。测试：输入：((5+3*(4+2)*12)+3)/(1+3)+5 输出：5 3 4 2 + * 12 * + 3 + 1 3 + / 5 +

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