PHP實現： 

復制代碼代碼如下:

<?php 
/** 
* 計算四則運算表達式 
*/ 
error_reporting(E_ALL); 
$exp = '(1+2*(3+5)/4)*(3+(5-4)*2)'; 
$arr_exp = array(); 
for($i=0;$i<strlen($exp);$i++){ 
$arr_exp[] = $exp[$i]; 
} 
$result = calcexp( array_reverse($arr_exp) ); 
echo $exp . '=' . $result; 
function calcexp( $exp ){ 
$arr_n = array(); 
$arr_op = array(); 
while( ($s = array_pop( $exp )) != '' ){ 
if( $s == '(' ){ 
$temp = array(); $quote = 1; $endquote = 0; 
while( ($t = array_pop($exp)) != '' ){ 
if( $t == '(' ){ 
$quote++; 
} 
if( $t == ')' ){ 
$endquote++; 
if( $quote == $endquote ){ 
break; 
} 
} 
array_push($temp, $t); 
} 
$temp = array_reverse($temp); 
array_push($arr_n, calcexp($temp) ); 
}else if( $s == '*' || $s == '/' ){ 
$n2 = array_pop($exp); 
if( $n2 == '(' ){ 
$temp = array(); $quote = 1; $endquote = 0; 
while( ($t = array_pop($exp)) != '' ){ 
if( $t == '(' ){ 
$quote++; 
} 
if( $t == ')' ){ 
$endquote++; 
if( $quote == $endquote ) 
break; 
} 
array_push($temp, $t); 
} 
$temp = array_reverse($temp); 
$n2 = calcexp($temp); 
} 
$op = $s; 
$n1 = array_pop($arr_n); 
$result = operation($n1, $op, $n2); 
array_push($arr_n, $result); 
}elseif( $s == '+' || $s == '-' ){ 
array_push($arr_op, $s); 
}else{ 
array_push($arr_n, $s); 
} 
} 
$n2 = array_pop($arr_n); 
while( ($op = array_pop($arr_op)) != '' ){ 
$n1 = array_pop($arr_n); 
$n2 = operation($n1, $op, $n2); 
} 
return $n2; 
} 
function operation( $n1, $op, $n2 ){ 
switch ($op) { 
case '+': 
return intval($n1) + intval($n2); 
break; 
case '-': 
return intval($n1) - intval($n2); 
break; 
case '*': 
return intval($n1) * intval($n2); 
break; 
case '/': 
return intval($n1) / intval($n2); 
break; 
} 
} 

這個實現方式中使用了兩個堆棧，一個用來存儲數字，一個用來存儲運算符，遇到括號以后就遞歸進入括號內運算，實現方式有點笨拙，后面補充一下“逆波蘭表達式”的算法實現。