From 0e1cc6d2771113ef0caeed9a26ec0e7e590e91e7 Mon Sep 17 00:00:00 2001
From: Meliodas417 <76930518+Meliodas417@users.noreply.github.com>
Date: Sun, 24 Mar 2024 11:07:42 -0500
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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除法逻辑不完整，假如是5/2=2没问题，但如果是-5/2,会向下取整，就变成-3了，可以单独定义一个函数。

底下最先弹出的数字实际上是最后进入的数字，也就是运算符右边的操作数，而第二个弹出的数字是第一个进入的数字，即运算符左边的操作数。stack.append(operation(op2, op1))
---
 problems/0150.逆波兰表达式求值.md | 41 ++++++++++++++++-------
 1 file changed, 29 insertions(+), 12 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 663a68ea..fd4d0726 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
@@ -169,8 +169,12 @@ class Solution {
 ```python
 from operator import add, sub, mul
 
-class Solution:
-    op_map = {'+': add, '-': sub, '*': mul, '/': lambda x, y: int(x / y)}
+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: List[str]) -> int:
         stack = []
@@ -186,18 +190,31 @@ class Solution:
 
 另一种可行，但因为使用eval相对较慢的方法:
 ```python
-class Solution:
-    def evalRPN(self, tokens: List[str]) -> int:
+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
+        """
         stack = []
-        for item in tokens:
-            if item not in {"+", "-", "*", "/"}:
-                stack.append(item)
+        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))
             else:
-                first_num, second_num = stack.pop(), stack.pop()
-                stack.append(
-                    int(eval(f'{second_num} {item} {first_num}'))   # 第一个出来的在运算符后面
-                )
-        return int(stack.pop()) # 如果一开始只有一个数，那么会是字符串形式的
+                stack.append(int(token))
+        return stack.pop()
+
 
 ```