[LintCode] Evaluate Reverse Polish Notation 计算逆波兰表达式

  

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Have you met this question in a real interview? 

Yes

Example

["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

 

LeetCode上的原题，请参见我之前的博客Evaluate Reverse Polish Notation。

 

解法一：

class Solution {
public:
    /**
     * @param tokens The Reverse Polish Notation
     * @return the value
     */
    int evalRPN(vector<string>& tokens) {

    stack<int> s;
        for (auto a : tokens) {
            if (a == "+" || a == "-" || a == "*" || a == "/") {
                if (s.size() < 2) break;
                int t = s.top(); s.pop();
                int k = s.top(); s.pop();
                if (a == "+") k += t;
                else if (a == "-") k -= t;
                else if (a == "*") k *= t;
                else if (a == "/") k /= t;
                s.push(k);
            } else {
                s.push(stoi(a));
            }
        }
        return s.top();
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param tokens The Reverse Polish Notation
     * @return the value
     */
    int evalRPN(vector<string>& tokens) {
        int op = tokens.size() - 1;
        return helper(tokens, op);
    }
    int helper(vector<string>& tokens, int& op) {
        string s = tokens[op];
        if (s == "+" || s == "-" || s == "*" || s == "/") {
            int v2 = helper(tokens, --op);
            int v1 = helper(tokens, --op);
            if (s == "+") return v1 + v2;
            else if (s == "-") return v1 - v2;
            else if (s == "*") return v1 * v2;
            else return v1 / v2;
        } else {
            return stoi(s);
        }
    }
};