Re: [閒聊] 每日LeetCode

[z2147483648 (溢傷了喇)]

※ 引述《JerryChungYC (JerryChung)》之銘言：
: https://leetcode.com/problems/evaluate-reverse-polish-notation
: 150. Evaluate Reverse Polish Notation
: 回傳逆波蘭表示法的結果值
: 有效運算子為 "+" "-" "*" "/"
: 每個操作數會是整數或是另一個表達式
: 兩個整數之間的除法是向零截斷
: 不會有被零除的情況
: 答案和中間計算都可以用32位元整數表示
: Example 1:
: Input: tokens = ["2","1","+","3","*"]
: Output: 9
: Explanation: ((2 + 1) * 3) = 9
: Example 2:
: Input: tokens = ["4","13","5","/","+"]
: Output: 6
: Explanation: (4 + (13 / 5)) = 6

耍廢太久，要回來乖乖找工作了

我也來寫個

C++ code

#include<unordered_set>
class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        unordered_set<string> op = {"+", "-", "*", "/"};
        stack<string> s;
        for (int i = 0; i < tokens.size(); ++i)
        {
            if (op.find(tokens[i]) == op.end()) s.push(tokens[i]);
            else
            {
                int num1 = stoi(s.top());
                s.pop();
                int num2 = stoi(s.top());
                s.pop();
                int ret = operate(num1, num2, tokens[i]);
                s.push(to_string(ret));
            }
        }
        return stoi(s.top());
    }
    int operate(int num1, int num2, string op)
    {
        if (op == "+") return num2+num1;
        else if (op == "-") return num2-num1;
        else if (op == "*") return num2*num1;
        else if (op == "/") return num2/num1;
        else return 0;
    }
};

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