Evaluate Reverse Polish Notation
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
一道混合的题,难度不高,是几个部分组成。
需要注意:
1.转换数字,负数的存在。
2.出栈时数字的顺序。
3.好像没什么了……
class Solution {
public:
int evalRPN(vector<string> &tokens) {
stack<int> sk;
int result;
for(int i = 0 ; i < tokens.size(); i++)
{
if(isop(tokens[i]))
{
int second = sk.top();
sk.pop();
int first = sk.top();
sk.pop();
int midresult;
if(tokens[i] == "+") midresult = first + second;
else if(tokens[i] == "-")midresult = first - second;
else if(tokens[i] == "*")midresult = first * second;
else if(tokens[i] == "/")midresult = first / second;
sk.push(midresult);
}
else
{
int tp = string2num(tokens[i]);
sk.push(tp);
}
}
return sk.top();
}
bool isop(string s)
{
if(s== "+" ||s== "-" ||s== "*" ||s== "/" )return 1;
else return 0;
}
int string2num(string s)
{
int re =0 ;
int flag = 0;
int i = 0;
if(s[0] == '-')
{
flag =1;
i++;
}
for( ; i < s.size();i++)
{
re = re*10;
re = re + s[i]-'0';
}
if(flag ==1)
re = 0 -re;
return re;
}
};
posted on 2014-03-24 16:33 pengyu2003 阅读(104) 评论(0) 收藏 举报
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