From 0fa443ce1a234704898630c188640f4512e9f7d1 Mon Sep 17 00:00:00 2001
From: Yuan Yuan <yyccphil@gmail.com>
Date: Thu, 26 Dec 2024 14:03:38 -0600
Subject: [PATCH 1/2] =?UTF-8?q?fix:=20150=E9=A2=98=E6=9B=B4=E6=AD=A3Python?=
 =?UTF-8?q?=E8=A7=A3=E6=B3=95=E4=B8=AD=E4=BD=BF=E7=94=A8eval()=E7=9A=84?=
 =?UTF-8?q?=E6=96=B9=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

此处原本提供的两个python解法是一样的，并无区别；更正为实际上真正使用eval()的方法。
---
 problems/0150.逆波兰表达式求值.md | 32 +++++++----------------
 1 file changed, 9 insertions(+), 23 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 7d4031d7..4ffe950b 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
@@ -188,34 +188,20 @@ class Solution(object):
         return stack.pop()
 ```
 
-另一种可行，但因为使用eval相对较慢的方法:
+另一种可行，但因为使用eval()相对较慢的方法:
 ```python
-from operator import add, sub, mul
-
-def div(x, y):
-    # 使用整数除法的向零取整方式
-    return int(x / y) if x * y > 0 else -(abs(x) // abs(y))
-
 class Solution(object):
-    op_map = {'+': add, '-': sub, '*': mul, '/': div}
-
-    def evalRPN(self, tokens):
-        """
-        :type tokens: List[str]
-        :rtype: int
-        """
+    def evalRPN(self, tokens: List[str]) -> int:
         stack = []
         for token in tokens:
-            if token in self.op_map:
-                op1 = stack.pop()
-                op2 = stack.pop()
-                operation = self.op_map[token]
-                stack.append(operation(op2, op1))
+            # 判断是否为数字，因为isdigit()不识别负数，故需要排除第一位的符号
+            if token.isdigit() or (len(token)>1 and token[1].isdigit()):
+                stack.append(token)
             else:
-                stack.append(int(token))
-        return stack.pop()
-
-
+                op2 = stack.pop()
+                op1 = stack.pop()
+                stack.append(str(int(eval(op1 + token + op2))))
+        return int(stack.pop())
 ```
 
 ### Go:

From 0638ba5ad88909b1463ccb43f55cb06f9bc88d09 Mon Sep 17 00:00:00 2001
From: Yuan Yuan <yyccphil@gmail.com>
Date: Thu, 26 Dec 2024 14:09:06 -0600
Subject: [PATCH 2/2] =?UTF-8?q?docs:=20150=E9=A2=98=E6=B7=BB=E5=8A=A0?=
 =?UTF-8?q?=E6=B3=A8=E9=87=8A?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0150.逆波兰表达式求值.md | 1 +
 1 file changed, 1 insertion(+)

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 4ffe950b..5fb28c29 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
@@ -200,6 +200,7 @@ class Solution(object):
             else:
                 op2 = stack.pop()
                 op1 = stack.pop()
+		# 由题意"The division always truncates toward zero"，所以使用int()可以天然取整
                 stack.append(str(int(eval(op1 + token + op2))))
         return int(stack.pop())
 ```