LeetCode-Evaluate Reverse Polish Notation

最新推荐文章于 2021-03-20 19:59:28 发布

所难

最新推荐文章于 2021-03-20 19:59:28 发布

阅读量301

点赞数

CC 4.0 BY-SA版权

分类专栏： Cracking LeetCode String Math Game Stack 文章标签： leetcode Evaluate Reverse Pol

本文链接：https://blog.youkuaiyun.com/HolyDominat/article/details/38226445

Cracking LeetCode 同时被 3 个专栏收录

152 篇文章

订阅专栏

Math Game

50 篇文章

订阅专栏

String

16 篇文章

订阅专栏

本文介绍了一种解析逆波兰表达式的算法实现，包括如何使用栈数据结构处理加、减、乘、除运算符，并提供了相应的代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

Code:

<span style="font-size:14px;">class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        const int length = tokens.size();
        stack<int> stk;
        for (int i = 0; i < length; ++i)
            if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
                int second = stk.top();
                stk.pop();
                int first = stk.top();
                stk.pop();
                if (tokens[i] == "+")
                    stk.push(first+second);
                else if (tokens[i] == "-")
                    stk.push(first-second);
                else if (tokens[i] == "*")
                    stk.push(first*second);
                else if (tokens[i] == "/")
                    stk.push(first/second);
            } else
                stk.push(atoi(tokens[i].c_str()));
        return stk.top();
    }
};</span>