标签：examples string exp express class push token div tor
题目：
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
思路：
这种逆波兰表达式，很明显有栈来求解。需要注意的一点，就是来一个新的符号的时候，将栈中的两个值取出来进行操作，再放回栈中
此时先取出的为num2,后取出的为num1,进行num1 (+-*/) num2 操作
实现：
class Solution {
public:
int evalRPN(vector<string>& tokens) {
if (tokens.size() == 0){
return 0;
}
stack<int> st;
for (int i=0; i < tokens.size(); ++i){
string s = tokens[i];
if ("+" == s || "-" == s || "*" == s || "/" == s){
if (st.size() < 2){
return 0;
}
int num2 = st.top();
st.pop();
int num1 = st.top();
st.pop();
int result = 0;
if("+" == s){
result = num1 + num2;
}
else if ("-" == s){
result = num1 - num2;
}
else if ("*" == s){
result = num1 * num2;
}
else if ("/" == s){
result = num1 / num2;
}
st.push(result);
}
else{
st.push(atoi(s.c_str()));
}
}
return st.top();
}
};
标签：examples string exp express class push token div tor
原文地址：http://www.cnblogs.com/zhengbiao/p/7296460.html
