2019-05-02 15:59:01 +08:00
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# LeetCode 第 150 号问题：逆波兰表达式求值
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> 本文首发于公众号「五分钟学算法」，是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
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>
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> 个人网站：[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
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题目来源于 LeetCode 上第 150 号问题：逆波兰表达式求值。题目难度为 Medium，目前通过率为 43.7% 。
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### 题目描述
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根据[逆波兰表示法](https://baike.baidu.com/item/%E9%80%86%E6%B3%A2%E5%85%B0%E5%BC%8F/128437)，求表达式的值。
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有效的运算符包括 `+`, `-`, `*`, `/` 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。
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**说明：**
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- 整数除法只保留整数部分。
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- 给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
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**示例 1：**
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```
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输入: ["2", "1", "+", "3", "*"]
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输出: 9
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解释: ((2 + 1) * 3) = 9
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```
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**示例 2：**
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```
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输入: ["4", "13", "5", "/", "+"]
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输出: 6
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解释: (4 + (13 / 5)) = 6
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```
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**示例 3：**
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```
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输入: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
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输出: 22
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解释:
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((10 * (6 / ((9 + 3) * -11))) + 17) + 5
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= ((10 * (6 / (12 * -11))) + 17) + 5
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= ((10 * (6 / -132)) + 17) + 5
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= ((10 * 0) + 17) + 5
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= (0 + 17) + 5
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= 17 + 5
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= 22
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```
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### 题目解析
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用数据结构`栈`来解决这个问题。
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- 从前往后遍历数组
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- 遇到数字则压入栈中
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- 遇到符号，则把栈顶的两个数字拿出来运算，把结果再压入栈中
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- 遍历完整个数组，栈顶数字即为最终答案
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### 动画描述
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2019-11-14 11:00:28 +08:00
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2019-05-02 15:59:01 +08:00
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### 代码实现
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```
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class Solution {
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public:
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int evalRPN(vector<string>& tokens) {
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stack<int> nums;
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stack<char> ops;
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for(const string& s: tokens){
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if(s == "+" || s == "-" || s == "*" || s == "/"){
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int a = nums.top();
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nums.pop();
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int b = nums.top();
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nums.pop();
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if(s == "+"){
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nums.push(b + a);
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}else if(s == "-"){
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nums.push(b - a);
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} else if(s == "*"){
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nums.push(b * a);
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}else if(s == "/"){
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nums.push(b / a);
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}
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}
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else{
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nums.push(atoi(s.c_str()));
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}
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}
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return nums.top();
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}
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};
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```
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2019-11-14 11:00:28 +08:00
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