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Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
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题意：求逆波兰表达式的结果
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public
class
Solution {
public
int
isopnd(String s){
if
(s.length()!=
1
)
return
0
;
char
c=s.charAt(
0
);
switch
(c){
case
'+'
:
return
1
;
case
'-'
:
return
2
;
case
'*'
:
return
3
;
case
'/'
:
return
4
;
default
:
return
0
;
}
}
public
int
evalRPN(String[] tokens) {
if
(tokens.length==
1
)
return
Integer.valueOf(tokens[
0
]);
Stack<Integer> stack=
new
Stack<Integer>();
int
ans=
0
;
for
(String tk : tokens){
int
op=isopnd(tk);
if
(op!=
0
){
int
a=stack.peek();
stack.pop();
int
b=stack.peek();
stack.pop();
if
(op==
1
){
ans=a+b;
}
if
(op==
2
){
ans=b-a;
}
if
(op==
3
){
ans=a*b;
}
if
(op==
4
){
ans=b/a;
}
System.out.println(ans);
stack.push(ans);
}
else
{
stack.push(Integer.valueOf(tk));
}
}
return
ans;
}
}
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PS：用栈。第一次调用栈。遇到数字，入栈，遇到运算符，出栈俩数字做计算，结果再入栈。'
本文转自 努力的C 51CTO博客，原文链接:http://blog.51cto.com/fulin0532/1902571
