算法Sedgewick第四版-第1章基础-007一用两个栈实现简单的编译器

1.

 1 package algorithms.util;
 2 
 3 import algorithms.ADT.Stack;
 4 
 5 /******************************************************************************
 6  *  Compilation:  javac Evaluate.java
 7  *  Execution:    java Evaluate
 8  *  Dependencies: Stack.java
 9  *
10  *  Evaluates (fully parenthesized) arithmetic expressions using
11  *  Dijkstra's two-stack algorithm.
12  *
13  *  % java Evaluate 
14  *  ( 1 + ( ( 2 + 3 ) * ( 4 * 5 ) ) ) 
15  *  101.0 
16  *
17  *  % java Evaulate
18  *  ( ( 1 + sqrt ( 5 ) ) / 2.0 ) 
19  *  1.618033988749895
20  *
21  *
22  *  Note: the operators, operands, and parentheses must be
23  *  separated by whitespace. Also, each operation must
24  *  be enclosed in parentheses. For example, you must write
25  *  ( 1 + ( 2 + 3 ) ) instead of ( 1 + 2 + 3 ).
26  *  See EvaluateDeluxe.java for a fancier version.
27  *
28  *
29  *  Remarkably, Dijkstra's algorithm computes the same
30  *  answer if we put each operator *after* its two operands
31  *  instead of *between* them.
32  *
33  *  % java Evaluate
34  *  ( 1 ( ( 2 3 + ) ( 4 5 * ) * ) + ) 
35  *  101.0
36  *
37  *  Moreover, in such expressions, all parentheses are redundant!
38  *  Removing them yields an expression known as a postfix expression.
39  *  1 2 3 + 4 5 * * + 
40  * 
41  *
42  ******************************************************************************/
43 
44 public class Evaluate {
45     public static void main(String[] args) { 
46         Stack<String> ops  = new Stack<String>();
47         Stack<Double> vals = new Stack<Double>();
48 
49         while (!StdIn.isEmpty()) {
50             String s = StdIn.readString();
51             if      (s.equals("("))               ;
52             else if (s.equals("+"))    ops.push(s);
53             else if (s.equals("-"))    ops.push(s);
54             else if (s.equals("*"))    ops.push(s);
55             else if (s.equals("/"))    ops.push(s);
56             else if (s.equals("sqrt")) ops.push(s);
57             else if (s.equals(")")) {
58                 String op = ops.pop();
59                 double v = vals.pop();
60                 if      (op.equals("+"))    v = vals.pop() + v;
61                 else if (op.equals("-"))    v = vals.pop() - v;
62                 else if (op.equals("*"))    v = vals.pop() * v;
63                 else if (op.equals("/"))    v = vals.pop() / v;
64                 else if (op.equals("sqrt")) v = Math.sqrt(v);
65                 vals.push(v);
66             }
67             else vals.push(Double.parseDouble(s));
68         }
69         StdOut.println(vals.pop());
70     }
71 }

 

posted @ 2016-04-18 21:00  shamgod  阅读(270)  评论(0)  编辑  收藏  举报
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