From 02fcd2c4a8f12782529b0fa787d3fd3f1f0fdca7 Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Fri, 20 Aug 2021 17:14:16 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200150.=E9=80=86=E6=B3=A2?=
 =?UTF-8?q?=E5=85=B0=E8=A1=A8=E8=BE=BE=E5=BC=8F=E6=B1=82=E5=80=BC.md=20pyt?=
 =?UTF-8?q?hon=E4=BB=A3=E7=A0=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

用format string提升效率，增加可读性，避免使用索引访问，直接使用切片。
---
 problems/0150.逆波兰表达式求值.md | 24 ++++++++++++-----------
 1 file changed, 13 insertions(+), 11 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 2b294337..bcde7d5b 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
@@ -223,17 +223,19 @@ var evalRPN = function(tokens) {
 python3
 
 ```python
-def evalRPN(tokens) -> int:
-    stack = list()
-    for i in range(len(tokens)):
-        if tokens[i] not in ["+", "-", "*", "/"]:
-            stack.append(tokens[i])
-        else:
-            tmp1 = stack.pop()
-            tmp2 = stack.pop()
-            res = eval(tmp2+tokens[i]+tmp1)
-            stack.append(str(int(res)))
-    return stack[-1]
+class Solution:
+    def evalRPN(self, tokens: List[str]) -> int:
+        stack = []
+        for item in tokens:
+            if item not in {"+", "-", "*", "/"}:
+                stack.append(item)
+            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
+
 ```