LeetCode - Evaluate Reverse Polish Notation

Evaluate Reverse Polish Notation

2014.2.25 23:42

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

Solution:

　　The Reverse Polish Notation is a postorder traversal of the syntax tree, which is constructed with operands and operators.

　　With a stack it would be easy to process the sequence and calculate the result.

　　Total time and space complexities are both O(n).

Accepted code:

// 1AC, simple training on stack operation.
#include <stack>
#include <vector>
using namespace std;

class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        int i, n;
        int op1, op2;
        stack<int> nums;
        bool is_op;
        
        n = (int)tokens.size();
        for (i = 0; i < n; ++i) {
            is_op = false;
            if (tokens[i].length() == 1) {
                switch(tokens[i][0]) {
                case '+':
                case '-':
                case '*':
                case '/':
                    is_op = true;
                    break;
                }
            }
            
            if (is_op) {
                if (nums.size() < 2) {
                    // not enough operands
                    return 0;
                }
                op2 = nums.top();
                nums.pop();
                op1 = nums.top();
                nums.pop();
                switch (tokens[i][0]) {
                case '+':
                    nums.push(op1 + op2);
                    break;
                case '-':
                    nums.push(op1 - op2);
                    break;
                case '*':
                    nums.push(op1 * op2);
                    break;
                case '/':
                    if (op2 == 0) {
                        // divide by 0
                        return 0;
                    }
                    nums.push(op1 / op2);
                    break;
                }
            } else {
                if (sscanf(tokens[i].c_str(), "%d", &op1) != 1) {
                    // invalid integer format
                    return 0;
                }
                nums.push(op1);
            }
        }
        int result = nums.top();
        nums.pop();
        
        return result;
    }
};