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POConverting_infix_expressions_to_postfix_expressions
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  1. POConverting_infix_expressions_to_postfix_expressions
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    copied!<p>I thought that this code should solve the problem with inputs with parentheses like </p> <pre><code>1 * 2 + 3 * ( 5 + 2 ) - 3 </code></pre> <p>and output the postfix expression without parentheses, but the output is </p> <pre><code>1.0 2.0 * 3.0 * + 5.0 2.0 + ( 3.0 - ) </code></pre> <p>and another example: </p> <pre><code>9 - 5 * 8 + 19 - ( 8 + 10 ) </code></pre> <p>will be: </p> <pre><code>9.0 5.0 8.0 * - 19.0 + - 8.0 10.0 + ( ) </code></pre> <p>So I want to know why this code can't handle this case. I want the return string without parentheses by a good manner without deleting the parentheses at the end of the expression. </p> <pre><code>public static String Converting_infix_expressions_to_postfix_expressions(String infix) throws Exception{ StringTokenizer st = new StringTokenizer(infix); int numOF_tokens = st.countTokens(); String postfix = "" ; for (int i = 1; i &lt;= numOF_tokens; i++) { String term = st.nextToken(); try { // if it is an Float there is no problem will happen float x = Float.parseFloat(term); postfix += x +" " ; } catch (Exception e) { if(stack.isEmpty()) stack.push(term); else if(term == "(") stack.push(term); else if(term == ")"){ while(!stack.isEmpty() &amp;&amp; (String)stack.peek() != "(") postfix += stack.pop() +" "; stack.pop(); } else{ int x = 0,y = 0; switch(term){ case "+": x = 1; break; case "-": x = 1; break; case "*": x = 2; break; case "/": x = 2; break; } switch((String)stack.peek()){ case "+": y = 1; break; case "-": y = 1; break; case "*": y = 2; break; case "/": y = 2; break; } if(x &gt; y) stack.push(term); else { int x1 = x , y1 = y; boolean puchedBefore = false; while(x1 &lt;= y1){ //when the operator in the peak of the stack is smaller than one read before from the infix excepression. postfix += stack.pop() +" "; if(stack.isEmpty() || stack.peek() == "(" ){ // to put the term in it's right place if it samller than all the previos operators until the begining of this subset excepression "if it is between parentheses or not " stack.push(term); puchedBefore = true; break; } else{ switch(term){ case "+": x1 = 1; break; case "-": x1 = 1; break; case "*": x1 = 2; break; case "/": x1 = 2; break; } switch((String)stack.peek()){ case "+": y1 = 1; break; case "-": y1 = 1; break; case "*": y1 = 2; break; case "/": y1 = 2; break; } } } if(!puchedBefore) stack.push(term); } } } } while (! stack.isEmpty ()) { postfix += stack.pop () + " "; } System.out.println ("The postfix expression is : " + postfix); return postfix; } </code></pre>
 

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