題目要求：有一個四則運算的字符串表達式，編寫一個函數，計算四則運算的結果
PHP實現：
1 <?php
2
3 /**
4 * 計算四則運算表達式
5 */
6
7 error_reporting(E_ALL);
8
9 $exp = '(1+2*(3+5)/4)*(3+(5-4)*2)';
10 $arr_exp = array();
11
12 for($i=0;$i<strlen($exp);$i++){
13 $arr_exp[] = $exp[$i];
14 }
15 $result = calcexp( array_reverse($arr_exp) );
16 echo $exp . '=' . $result;
17
18 function calcexp( $exp ){
19 $arr_n = array();
20 $arr_op = array();
21
22 while( ($s = array_pop( $exp )) != '' ){
23 if( $s == '(' ){
24 $temp = array(); $quote = 1; $endquote = 0;
25 while( ($t = array_pop($exp)) != '' ){
26 if( $t == '(' ){
27 $quote++;
28 }
29 if( $t == ')' ){
30 $endquote++;
31 if( $quote == $endquote ){
32 break;
33 }
34 }
35 array_push($temp, $t);
36 }
37 $temp = array_reverse($temp);
38 array_push($arr_n, calcexp($temp) );
39 }else if( $s == '*' || $s == '/' ){
40 $n2 = array_pop($exp);
41 if( $n2 == '(' ){
42 $temp = array(); $quote = 1; $endquote = 0;
43 while( ($t = array_pop($exp)) != '' ){
44 if( $t == '(' ){
45 $quote++;
46 }
47 if( $t == ')' ){
48 $endquote++;
49 if( $quote == $endquote )
50 break;
51 }
52 array_push($temp, $t);
53 }
54 $temp = array_reverse($temp);
55 $n2 = calcexp($temp);
56 }
57
58 $op = $s;
59 $n1 = array_pop($arr_n);
60
61 $result = operation($n1, $op, $n2);
62 array_push($arr_n, $result);
63 }elseif( $s == '+' || $s == '-' ){
64 array_push($arr_op, $s);
65 }else{
66 array_push($arr_n, $s);
67 }
68 }
69
70 $n2 = array_pop($arr_n);
71 while( ($op = array_pop($arr_op)) != '' ){
72 $n1 = array_pop($arr_n);
73 $n2 = operation($n1, $op, $n2);
74 }
75
76 return $n2;
77 }
78
79 function operation( $n1, $op, $n2 ){
80 switch ($op) {
81 case '+':
82 return intval($n1) + intval($n2);
83 break;
84 case '-':
85 return intval($n1) - intval($n2);
86 break;
87 case '*':
88 return intval($n1) * intval($n2);
89 break;
90 case '/':
91 return intval($n1) / intval($n2);
92 break;
93 }
94 }
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，遇到括號以後就遞歸進入括號內運算，實現方式有點笨拙，後面補充一下“逆波蘭表達式”的算法實現。
