表达式运算，网上查了一下java有源代码。不过问题太多。而且都是一个版本。
根本没法使用。
逆波兰式算法主要的设计思路是把一般的中序表达式变为右序表达式
例子1+2-5*(5-4)*6-(6-1)
把上面的表达式转化为：
12+554-*6*-61–
然后通过栈再来计算。
具体的理论网上找找逆波兰式
根据人家的版本进行了修改。
package
expression1;
import
java.util.
*
;
//
栈类
public
class
Stacks
…
{
private
LinkedList list
=
new
LinkedList();
int
top
=-
1
;
public
void
push(Object value)
…
{
top
++
;
list.addFirst(value);
}
public
Object pop()
…
{
Object temp
=
list.getFirst();
top
—
;
list.removeFirst();
return
temp;
}
public
Object top()
…
{
return
list.getFirst();
}
}
package
expression1;
import
java.io.
*
;
import
java.util.
*
;
public
class
Expression
…
{
private
ArrayList expression
=
new
ArrayList();
//
存储中序表达式
private
ArrayList right
=
new
ArrayList();
//
存储右序表达式
private
String result;
//
结果
//
依据输入信息创建对象，将数值与操作符放入ArrayList中
private
Expression(String input)
…
{
StringTokenizer st
=
new
StringTokenizer(input,
”
+-*/()
”
,
true
);
while
(st.hasMoreElements())
…
{
String s
=
st.nextToken();
expression.add(s);
}
}
//
将中序表达式转换为右序表达式
private
void
toRight()
…
{
Stacks aStack
=
new
Stacks();
String operator;
int
position
=
0
;
while
(
true
)
…
{
if
(Calculate.isOperator((String) expression.get(position)))
…
{
if
(aStack.top
==
–
1
||
((String) expression.get(position)).equals(
”
(
”
))
…
{
aStack.push(expression.get(position));
}
else
…
{
if
(((String) expression.get(position)).equals(
”
)
”
))
…
{
while
(
true
)
…
{
if
(aStack.top
!=
–
1
&&!
((String) aStack.top()).equals(
”
(
”
))
…
{
operator
=
(String) aStack.pop();
right.add(operator);
}
else
…
{
if
(aStack.top
!=
–
1
)
aStack.pop();
break
;
}
}
}
else
…
{
while
(
true
)
…
{
if
(aStack.top
!=
–
1
&&
Calculate.priority((String) expression
.get(position))
<=
Calculate
.priority((String) aStack.top())
)
…
{
operator
=
(String) aStack.pop();
if
(
!
operator.equals(
”
(
”
))
right.add(operator);
}
else
…
{
break
;
}
}
aStack.push(expression.get(position));
}
}
}
else
right.add(expression.get(position));
position
++
;
if
(position
>=
expression.size())
break
;
}
while
(aStack.top
!=
–
1
)
…
{
operator
=
(String) aStack.pop();
if
(
!
operator.equals(
”
(
”
))
right.add(operator);
}
}
//
对右序表达式进行求值
private
void
getResult()
…
{
this
.toRight();
for
(
int
i
=
0
;i
<
right.size();i
++
)
…
{
System.out.println(right.get(i));
}
Stacks aStack
=
new
Stacks();
String op1, op2, is
=
null
;
Iterator it
=
right.iterator();
while
(it.hasNext())
…
{
is
=
(String) it.next();
if
(Calculate.isOperator(is))
…
{
op1
=
(String) aStack.pop();
op2
=
(String) aStack.pop();
aStack.push(Calculate.twoResult(is, op1, op2));
}
else
aStack.push(is);
}
result
=
(String) aStack.pop();
it
=
expression.iterator();
while
(it.hasNext())
…
{
System.out.print((String) it.next());
}
System.out.println(
”
=
”
+
result);
}
public
static
void
main(String avg[])
…
{
try
…
{
System.out.println(
”
Input a expression:
”
);
BufferedReader is
=
new
BufferedReader(
new
InputStreamReader(
System.in));
for
(;;)
…
{
String input
=
new
String();
input
=
is.readLine().trim();
if
(input.equals(
”
q
”
))
break
;
else
…
{
Expression boya
=
new
Expression(input);
boya.getResult();
}
System.out
.println(
”
Input another expression or input ‘q’ to quit:
”
);
}
is.close();
}
catch
(IOException e)
…
{
System.out.println(
”
Wrong input!!!
”
);
}
}
}
ackage expression1;
public
class
Calculate
…
{
//
判断是否为操作符号
public
static
boolean
isOperator(String operator)
…
{
if
(operator.equals(
”
+
”
)
||
operator.equals(
”
–
”
)
||
operator.equals(
”
*
”
)
||
operator.equals(
”
/
”
)
||
operator.equals(
”
(
”
)
||
operator.equals(
”
)
”
))
return
true
;
else
return
false
;
}
//
设置操作符号的优先级别
public
static
int
priority(String operator)
…
{
if
(operator.equals(
”
+
”
)
||
operator.equals(
”
–
”
))
return
1
;
else
if
(operator.equals(
”
*
”
)
||
operator.equals(
”
/
”
))
return
2
;
else
return
0
;
}
//
做2值之间的计算
public
static
String twoResult(String operator, String a, String b)
…
{
try
…
{
String op
=
operator;
String rs
=
new
String();
double
x
=
Double.parseDouble(b);
double
y
=
Double.parseDouble(a);
double
z
=
0
;
if
(op.equals(
”
+
”
))
z
=
x
+
y;
else
if
(op.equals(
”
–
”
))
z
=
x
–
y;
else
if
(op.equals(
”
*
”
))
z
=
x
*
y;
else
if
(op.equals(
”
/
”
))
z
=
x
/
y;
else
z
=
0
;
return
rs
+
z;
}
catch
(NumberFormatException e)
…
{
System.out.println(
”
input has something wrong!
”
);
return
”
Error
”
;
}
}
}
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