Question

我正在尝试从arraylist中读取符号，然后对它之前和之后的两个数字进行表达式（例如i-1 * i + 1）。但是，生成的结果始终是5、6或12！我真的找不到代码中的问题所在。

我之所以使用arraylist是因为它比数组更具动态性，从某种意义上说，我可以删除元素而不必关心'array'的长度。

    String replaceStr = replacer.replace("+", " + ");

我正在尝试添加空间以将整数用作元素，而不是使用charAt，这将不允许我将大于一位数的数字带入方程式。

    Double Computing(String equation) {
    double result = 0.0;
    double multiplResult = 1.0; //

    String replacer = equation;

    String replaceStr = replacer.replace("+", " + ");
    String replaceStr1 = replaceStr.replace("x", " x ");
    String replaceStr2 = replaceStr1.replace("-", " - ");
    String replaceStr3 = replaceStr2.replace("/", " / ");
    String[] items = replaceStr3.split("\\s*");

    ArrayList<String> list = new ArrayList<String>((Arrays.asList(items)));




   for (int i = 0; i < list.size(); i++ ) {
       String NewNum;

       if (list.get(i) == "x" || list.get(i) == "/"){

           if(list.get(i) == "x") {
               NewNum = String.valueOf(Double.parseDouble(list.get(i - 1)) * Double.parseDouble(list.get(i + 1)));
               list.set(i, NewNum);
               list.remove(i-1);
               list.remove(i+1);
           }
           else if (list.get(i) == "/"){
               NewNum = String.valueOf(Double.parseDouble(list.get(i - 1)) / Double.parseDouble(list.get(i + 1)));
               list.set(i, NewNum);
               list.remove(i-1);
               list.remove(i+1);
           }
           multiplResult *= Double.parseDouble(list.get(0));
       }
       if (list.get(i) == "+" || list.get(i) == "-"){

           if(list.get(i) == "+") {
               NewNum = String.valueOf(Double.parseDouble(list.get(i - 1)) + Double.parseDouble(list.get(i + 1)));
               list.set(i, NewNum);
               list.remove(i-1);
               list.remove(i+1);
           }
           else if (list.get(i) == "-"){
               NewNum = String.valueOf(Double.parseDouble(list.get(i - 1)) - Double.parseDouble(list.get(i + 1)));
               list.set(i, NewNum);
               list.remove(i-1);
               list.remove(i+1);
           }
           multiplResult *= Double.parseDouble(list.get(0));
       }
       result += multiplResult;
    }
return result;
}

Answer 1

如果仅使用+，-，x和/且不使用方括号，并且等式的语法正确，则可以这样做。但是您必须先评估x和/：

Double Computing(String equation) {
    String replace = equation.replace("+", " + ").replace("x", " x ").replace("-", " - ").replace("/", " / ");
    List<Object> list = new ArrayList<>(Arrays.asList(replace.split("\\s+")));

    for (int i = 0; i < list.size(); i++) {
        try {
            list.set(i, Double.parseDouble(list.get(i).toString()));
        } catch (NumberFormatException e) {
        }
    }

    while (list.contains("x")) {
        int i = list.indexOf("x");
        double mult = (Double) list.get(i - 1) * (Double) list.get(i + 1);
        list.remove(i + 1);
        list.set(i, mult);
        list.remove(i - 1);
    }
    while (list.contains("/")) {
        int i = list.indexOf("/");
        double div = (Double) list.get(i - 1) / (Double) list.get(i + 1);
        list.remove(i + 1);
        list.set(i, div);
        list.remove(i - 1);
    }
    double sum = (Double) list.get(0);
    for (int i = 1; i < (list.size() - 1); i += 2) {
        if (list.get(i).equals("+")) {
            sum += (Double) list.get(i + 1);
        } else {
            sum -= (Double) list.get(i + 1);
        }
    }

    return sum;
}

Answer 2

这是一个简单的评估器：

public class Evaluator {
    private final char[] chars;
    private int pos;

    public static double evaluate(String expr) {
        return new Evaluator(expr).expression();
    }

    private Evaluator(String expr) {
        chars = expr.toCharArray();
    }

    private double expression() {
        double value = term();
        char op;
        while (skipSpaces() && ((op = chars[pos]) == '+' || op == '-')) {
            ++pos;
            double other = term();
            switch (op) {
                case '+':
                    value += other;
                    break;
                case '-':
                    value -= other;
                    break;
            }
        }
        return value;
    }

    private double term() {
        double value = factor();
        char op;
        while (skipSpaces() && ((op = chars[pos]) == 'x' || op == '/')) {
            ++pos;
            double other = factor();
            switch (op) {
                case 'x':
                    value *= other;
                    break;
                case '/':
                    value /= other;
                    break;
            }
        }
        return value;
    }

    private double factor() {
        if (skipSpaces() && chars[pos] == '(') {
            ++pos;
            double result = expression();
            if (skipSpaces() && chars[pos] == ')') {
                ++pos;
            }
            return result;
        }
        return number();
    }

    private double number() {
        if (!skipSpaces()) {
            return 0;
        }
        int start = pos;
        if (chars[pos] == '+' || chars[pos] == '-') {
            ++pos;
        }
        if (pos >= chars.length || !Character.isDigit(chars[pos])) {
            return 0;
        }
        do {
            ++pos;
        } while (pos < chars.length
                && (Character.isDigit(chars[pos]) || chars[pos] == '.'));
        return Double.parseDouble(new String(chars, start, pos-start));
    }

    private boolean skipSpaces() {
        while (pos < chars.length && Character.isWhitespace(chars[pos])) {
            ++pos;
        }
        return pos < chars.length;
    }
}

如何对字符串数组列表中的两个数字进行数学表达

2 个答案: