From b0ec03922ea2081f0aa72e7fccf8a322aa9a9efb Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= <sunxiuyang04@gmail.com>
Date: Sun, 2 Oct 2022 10:46:19 +0800
Subject: [PATCH] =?UTF-8?q?Update=200150.=E9=80=86=E6=B3=A2=E5=85=B0?=
 =?UTF-8?q?=E8=A1=A8=E8=BE=BE=E5=BC=8F=E6=B1=82=E5=80=BC.md?=
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力扣修改了后端测试数据
---
 problems/0150.逆波兰表达式求值.md | 11 ++++-------
 1 file changed, 4 insertions(+), 7 deletions(-)

diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md
index 257bbb9a..a0ec5db3 100644
--- a/problems/0150.逆波兰表达式求值.md
+++ b/problems/0150.逆波兰表达式求值.md
@@ -89,10 +89,7 @@ C++代码如下：
 class Solution {
 public:
     int evalRPN(vector<string>& tokens) {
-        // 考虑到第20个样例使用int会溢出
-        // 此处使用long long来存储number
-        // 在最后用int()强行转换成int输出
-        
+        // 力扣修改了后台测试数据，需要用longlong
         stack<long long> st; 
         for (int i = 0; i < tokens.size(); i++) {
             if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
@@ -105,13 +102,13 @@ public:
                 if (tokens[i] == "*") st.push(num2 * num1);
                 if (tokens[i] == "/") st.push(num2 / num1);
             } else {
-                st.push(atoll(tokens[i].c_str()));
+                st.push(stoll(tokens[i]));
             }
         }
 
-        long long result = st.top();
+        int result = st.top();
         st.pop(); // 把栈里最后一个元素弹出（其实不弹出也没事）
-        return int(result);
+        return result;
     }
 };