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..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
@@ -188,34 +188,21 @@ 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()
+		# 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())
 ```
 
 ### Go: