Question

我有一个接收二进制表达式树的函数，并返回一个带有表达式的String。唯一的“问题”是结果表达式括号太多，
例如：该函数返回（a +（b * c）），但可以减少到+ b * c 它由二元运算符+， - ，*，/和一元运算符_（负数）定义我真正想知道的是，我是否可以修改现有函数以有效减少括号数，或创建另一个使用有序表达式字符串操作的函数。
功能如下：

private static String infijo(ArbolB t){
    String s = "";
    if (t != null) {
        String info = String.valueOf(t.info);
        if ("+-*/".contains(info)) s += "(";
        if ("_".contains(info)) s += "-(";
        s += infijo(t.left) + (info.equals("_") ? "" : info) + infijo(t.right);
        if ("+-*/_".contains(String.valueOf(t.info))) s += ")";
    }
    return s;
}

其中ArbolB是由以下内容定义的二叉树：

public class ArbolB {

    ArbolB right;
    ArbolB left;
    Object info;

    public ArbolB(Object info, ArbolB right, ArbolB left){
        this.info = info;
        this.right = right;
        this.left = left;
    }
}

Answer 1

写完这一切之后，我意识到我没有正确回答你的问题（我的解决方案忽略了PEMDAS，只是匹配对，d＆＃39;哦！）。所以，从中得到你能做到的......我不会把它丢掉：P

我认为你可以解决这个问题，但是这将是我首选的方法，使用和信任你已经拥有的东西。这可能是使用节点执行此操作的好方法，但为什么不使用您拥有的，对吧？

从您将表达式作为字符串开始（例如＆＃34;（（2 * 2）+ _（3 + 3））＆＃34;您可以尝试以下内容：

public string RemoveUnnecessaryParens(string expression)
{
    readonly string openParen = "(";
    readonly string closeParen = ")";
    // char array would also work for this
    // multiple ways to track "balance" of parens, lets use int
    int unclosedParenCount = 0;

    string toReturn = "";

    // iterate through the expression 
    for (int i = 0; i < expression.Length; i++)
    {
        string current = expression.Substring(i,1);
        if (openParen == current)
            unclosedParenCount++;
        else if (closedParen == current)
            unclosedParenCount--;
        else
            toReturn += current;

        if (unclosedParenCount < 0) throw new UnbalancedExpressionException(); // One more close than open! Uh oh!
    }

    if (0 != unclosedParenCount) throw new UnbalancedExpressionException(); // One more open than close at the end! Uh oh!
    else return toReturn;
}

有意义吗？

Answer 2

好吧，经过一段时间的考虑，我自己找到了一个解决方案，通过添加优先级函数来确定何时需要括号，以及一个变量来指示操作是在公式的左侧还是右侧，这是因为a-b + c不需要括号，但c +（ab）确实需要括号。

private static String infijo(ArbolB t, int priority, boolean right) {
    String s = "";
    int oP = 0;
    if (t != null) {
        String info = String.valueOf(t.info);
        int pi = priority(info);
        if ("+-*/".contains(info)) {
            /* this only adds parentheses if the operation has higher priority or if the 
               operation on the right side should be done before the one on the left side*/
            if ("+-*/".contains(info)) {
                if (pi/2 < priority/2) s += "(";
                else s += pi/2 == priority/2 && pi != priority && right ? "(" : "";
                oP = priority; //stores the old priority
                priority= pi; //priority of the new operator
            }
        }
        if ("_".contains(info)) {
            s += "-";
            oP = priority;
            priority = pi;
        }
        s += infijo(t.left, priority, false) + (info.equals("_") ? "" : info) 
              + infijo(t.right, priority, true);
        if ("+-*/".contains(info)) {
         // adds the closing parentheses following the same rules as for the opening ones
            if (priority / 2 < oP / 2) s += ")";
            else s += priority / 2 == oP / 2 && priority != oP && right ? ")" : "";
        }
    }
    return s;
}

private static int priority(String op) {
    if ("_".contains(op))   return 4;
    if ("/".contains(op))   return 3;
    if ("*".contains(op))   return 2;
    if ("-".contains(op))   return 1;
                            return 0;
}

@Override
public String toString() {
    ArbolB f = getFormula(); //this returns the Binary Expression Tree
    return infijo(f, Integer.MIN_VALUE, false);
}

减少二进制表达式树

2 个答案: