Postfix to Infix with minimum number of parentheses

问题

I am looking for algorithm postfix to infix notation which will produce the minimum number of the parentheses.

I have found that but it will produce many, many parentheses: http://tajendrasengar.blogspot.com/2011/09/postfix-to-infix-algorithm.html

For example

The input:

<ONP>abcd*/+~

The result:

<INF>~(a+b/(c*d))

回答1:

What you need to do if you really want as few parentheses as possible, is similar to what the algorithm you linked to says. However...

You should store an operator for each composite operand in the Stack. Namely, the last operator used in the operand. You could use a second Stack for this. If the operand is not composite, you could add null to the second Stack, since there is no operator.
Don't encapsulate the resulted String with parentheses. That is done elsewhere in the algorithm (see below).

When you pop the top two values from each of the Stacks, you have 3 operators at hand:

The current operator
The last used operator in the first operand (if the operator exists)
The last used operator in the second operand (if the operator exists)

Depending on these three operators, you should encapsulate the first and/or second operand with parentheses, before combining them.

You could use operator precedence to determine whether there should be parentheses. The order goes like this: (none), {"*", "/"}, {"+", "-"}

The first operand needs parentheses if and only if its operator has a lower precedence than the current operator.
The second operand needs parentheses if its operator has a lower precedence than the current operator, or if they have equal precedence where the current operator is either "/" or "-".

The rest should be done the way your algorithm described.

回答2:

Here is the implementation:

public class PostfixToInfixConverter implements ExpressionConverter {

@Override
public String convert(String postfix) {
    String[] expressions = postfix.split("\\s");
     Deque<Expression> stack = new LinkedList<Expression>();
    for (String exp : expressions) {
        if (exp.equals("+") || exp.equals("-")) {
            String right = stack.pop().getExpression();
            String left = stack.pop().getExpression();
            Expression newExp = new Expression(right + exp + left, exp);
            stack.push(newExp);
        } else if (exp.equals("*") || exp.equals("/")) {
            String right = correctExpression(stack.pop());
            String left = correctExpression(stack.pop());
            stack.push(new Expression(right +  exp +  left, exp));
        } else {
            stack.push(new Expression(exp, ""));
        }
    }
    return stack.pop().getExpression();
}

private String correctExpression(Expression exp) {
    String result = exp.getExpression();
    if (exp.getOperatorUsed().equals("+") || exp.getOperatorUsed().equals("-")) {
        result = "(" + result + ")";
    }
    return result;
}

private static class Expression {
    String expression;
    String operatorUsed;
    public Expression(String exp, String operator) {
        this.expression = exp;
        this.operatorUsed = operator;
    }
    public String getExpression() {
        return expression;
    }
    public String getOperatorUsed() {
        return operatorUsed;
    }       
}

}

Here is the unit tests

@Test
public void test() {
    ExpressionConverter converter = new PostfixToInfixConverter();
    assertThat(converter.convert("1 2 * 3 4 * + 5 *"), equalTo("5*(4*3+2*1)"));
    assertThat(converter.convert("a b + c + 2 *"), equalTo("2*(c+b+a)"));
}

来源：https://stackoverflow.com/questions/15780817/postfix-to-infix-with-minimum-number-of-parentheses

标签

java

infix-notation

postfix-notation