Java RPN (Обратная польская запись) инфикс для добавления постфикса

Question

Java RPN (Обратная польская запись) инфикс для добавления постфикса

Я однажды записал грубой силе поиск ключа RC5, который обработал два ключа за один раз, первый ключ использовал целочисленный конвейер, второй ключ использовал конвейеры SSE, и эти два были чередованы на уровне инструкции. Это было тогда вместе с программой супервизора, которая выполнила экземпляр кода каждого ядра в системе. Всего, код работал приблизительно в 25 раз быстрее, чем наивная версия C.

5

java algorithm

задан Margus 24 August 2009 в 10:13

2 ответа

Not an exact answer to the specific question but something I'd recommend for developing these kinds of algorithms: have a look at test driven devlopment (TDD). In brief: write a couple of unit tests - for example with JUnit - for the infix2 method, where you feed the method with test patterns (expressions) and test, if infix2 produces the right output.

Start with easy ones, like

assertequals("1", "1"); // positive number
assertequals("-1", "-1"); // negative number
assertequals("1+1", "1 1 +"); // simple addition
assertequals(" 1 + 1 ", "1 1 +"); // simple addition with whitechars
assertequals(" 1 + +1 ", "1 -1 +"); // simple addition with pos. number & whitechars
assertequals(" 1 + -1 ", "1 -1 +"); // simple addition with neg. number & whitechars
assertequals("(1+1)", "1 1 +"); // simple addition with brackets

and don't forget illegal expressions like

String[] illegalExpressions = {null, "", " ", "1 +", "1 + 1)"};

The test cases for you examples should be

assertequals("52+(1+2)*4-3", "52 1 2 + 4 * 3 -");
assertequals("52+((1+2)*4)-3", "52 1 2 + 4 * 3 -");
assertequals("(52+1+2)*4-3", "52 1 + 2 + 4 * 3 -");

2

ответ дан 18 December 2019 в 11:59

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score 13 · Accepted Answer

The algorithm is pretty simple (and here is a good explanation). Every operation has a binding weight, with + and - being the lowest. There are two rules:

print out numbers immediately
never put a lighter item on a heavier item
left parentheses go on the stack
right parentheses pop off the stack until you hit a left parentheses, and then remove the left parentheses

Given your first example, 52+(1+2)*4-3, here is the stack:

 52+          => +
 52+(         => + (
 52+(1+       => + ( + 
 52+(1+2)     => +       //right parentheses popped +
 52+(1+2)*4   => + * 
 52+(1+2)*4-3 => + -     //can't put - on top of *, so pop off *
 ... and then pop the stack until it's empty.

Replacing your switch loop with the following (closest analog to what you had) will give correct answers for your three examples. In a real parser you would give each operator a weight and generalize the pop mechanism.

for (int i = 0; i < in.length; i++)
        switch (in[i]) {
        case '+':
        case '-':
            while (!stack.empty() && (stack.peek() == '*' || stack.peek() == '/')) {
                out.append(' ');
                out.append(stack.pop());
            }
            out.append(' ');
            stack.push(in[i]);
            break;
        case '*':
        case '/':
            out.append(' ');
            stack.push(in[i]);
            break;
        case '(':
            stack.push(in[i]);
            break;
        case ')':
            while (!stack.empty() && stack.peek() != '(') {
                out.append(' ');
                out.append(stack.pop());
            }
            stack.pop();
            break;
        default:
            out.append(in[i]);
            break;
        }