栈练习--Dijkstra的双栈算术表达式求值算法

Posted 2020-10-22 dreamy

此法还有待完善，比如在做除法运算时未判断除数是否为0，而且也可以加以扩展，比如log、sin、操作符优先级。。

 1 public class Evaluate {
 2     
 3     public static void main(String[] args) {
 4         String steam = "( 1 + ( 2 + 3 ) * ( 4 * 5 ) ) )";
 5 //        String steam = "( ( 1 + sqrt ( 5.0 ) ) / 2.0 )";
 6         Stack<String> ops = new Stack<String>();
 7         Stack<Double> vals = new Stack<Double>();
 8         
 9         String[] steams = steam.split(" ");
10         
11         for(int i=0; i<steams.length; i++) {
12             //读取字符，如果是运算符就压入栈
13             String s = steams[i];
14             /*char item = steam.charAt(i);
15             String s = String.valueOf(item);*/
16             
17             
18             if(s.equals("(")) {
19                 ;
20             }else if(s.equals("+")) {
21                 ops.push(s);
22             }else if(s.equals("-")) {
23                 ops.push(s);
24             }else if(s.equals("*")) {
25                 ops.push(s);
26             }else if(s.equals("/")) {
27                 ops.push(s);
28             }else if(s.equals("sqrt")) {
29                 ops.push(s);
30             }else if(s.equals(")")) {
31                 //如果字符为“）”则弹出运算符和操作数，计算结果并压入栈
32                 String op = ops.pop();
33                 double v = vals.pop();
34                 if(op.equals("+")) {
35                     v = vals.pop() + v;
36                 }else if(op.equals("-")) {
37                     v = vals.pop() - v;
38                 }else if(op.equals("*")) {
39                     v = vals.pop() * v;
40                 }else if(op.equals("/")) {
41                     v = vals.pop() / v;
42                 }else if(op.equals("sqrt")) {
43                     v = Math.sqrt(v);
44                 }
45                 vals.push(v);
46             }else {
47                 //如果字符既非运算符也不是括号，将它作为double值压入栈
48                 vals.push(Double.parseDouble(s));
49             }
50         }
51         System.out.println(vals.pop());
52     }
53 }