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
class Solution {
public:
bool isFlag(string str) {
if(str.length() != 1)
return false;
if(str[0] == '-' || str[0] == '+' || str[0] == '/' || str[0] == '*')
return true;
return false;
}
int calculate(int left, int right, string flag) {
if(flag[0] == '+')
return left + right;
if(flag[0] == '-')
return left - right;
if(flag[0] == '*')
return left * right;
if(flag[0] == '/')
return left / right;
return 0;
}
int stringToNumber(string str) {
int number = 0;
int flag = 1;
for(int ii = 0; ii < str.length(); ii ++) {
if(ii == 0) {
if(str[0] == '-') {
flag = -1;
continue;
}
else if(str[0] == '+') {
continue;
}
}
number = number * 10 + str[ii] - '0';
}
return number * flag;
}
int evalRPN(vector<string> &tokens) {
if(tokens.size() == 0)
return 0;
stack<int> st;
for(int ii = 0; ii < tokens.size(); ii ++) {
if(isFlag(tokens[ii])) {
int right = st.top();
st.pop();
int left = st.top();
st.pop();
int current = calculate(left, right, tokens[ii]);
st.push(current);
}
else {
st.push(stringToNumber(tokens[ii]));
}
}
return st.top();
}
};
本文介绍了一种通过逆波兰表达式(Reverse Polish Notation, RPN)进行算术运算的方法,并提供了一个C++实现的例子。该算法使用栈来处理RPN表达式中的操作数和运算符,支持加、减、乘、除四种基本运算。
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