【LeetCode OJ】Evaluate Reverse Polish Notation

Problem link:

http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/

According to the wiki page, the algorithm for evaluating any postfix expression is fairly straightforward. The following is the pseudo-code for evaluating a postfix expression.

while there are input tokens left
read the next token from input
if the token is a value (operand)
push it into the stack
else (the token is an operator)
let n be the number of the arguments of the operator
if there are fewer than n values in the stack
return ERROR (non-sufficient operands in the expression)
else
pop the top n values from the stack
evaluate the expression with the operator and operands
push the result, if any, back into the stack
if there is only one value in the stack
return the value as the result
else
return ERROR (too many operands in the expression)

However, in this problem, it is much easier since there are only four operators. What we need to concern is only to identify the operators and numbers (integer in this case). Therefore, we will implement two functions:

A stack structure

Functions on a string that determine the string is an integer or an operator.

The C++ implementation of the algorithm is as follows.

#include <stack>
#include <ctype.h>

class Solution {
public:
int evalRPN(vector<string> &tokens) {
// The stack to store operators (int values) only
std::stack<int> res;
int x,y;
// Iterate the vector from left to right
for(std::vector<string>::iterator it = tokens.begin(); it != tokens.end(); it++)
{
// If operator, pop and evaluate
if ( (*it).length() == 1 && !isdigit((*it)[0]) ) {
y = res.top();
res.pop();
x = res.top();
res.pop();
if (*it == "+") x += y;
else if (*it == "-") x -= y;
else if (*it == "*") x *= y;
else if (*it == "/") x /= y;
res.push(x);
}
// If operand, push it into the stack
else res.push( atoi((*it).c_str()) );
}
// The only value left in the stack is the final result
return res.top();
}
};