From 0cf3fd104b1068a3fbb026a5df9a56ee28b2838c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Wed, 21 Aug 2019 22:12:06 +0800
Subject: [PATCH] Create 150. Evaluate Reverse Polish Notation.md

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+# [150. Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation/)
+# 思路
+当时学栈的时候应该有学过逆波兰表达式, 逆波兰表达式就是把操作数放前面，把操作符后置的一种写法. 由于题目说了给的表达式肯定是合法的,
+所以我们可以很容易用栈解决: 从前往后遍历一遍表达式, 若是数字则进栈, 若是四则符号之一, 则pop出栈顶的两个元素进行相应运算, 并将运算结果压入栈中. 
+遍历完成后栈中只剩下一个元素即最终结果. 
+
+空间和时间复杂度都是O(n)
+
+注意我们可以使用函数`stoi()`很方便将字符串转换成整数. 
+
+# C++
+``` C++
+class Solution {
+public:
+    int evalRPN(vector<string>& tokens) {
+        stack<int> stk;
+        int i = 0;
+        for(int i = 0; i < tokens.size(); i++){
+            if(tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/") 
+                stk.push(stoi(tokens[i]));
+            else{
+                int b = stk.top(); stk.pop();
+                int a = stk.top(); stk.pop();
+                if(tokens[i] == "+") stk.push(a + b);
+                else if(tokens[i] == "-") stk.push(a - b);
+                else if(tokens[i] == "*") stk.push(a * b);
+                else if(tokens[i] == "/") stk.push(a / b);
+            }
+        }
+        return stk.top();
+    }
+};
+```