Merge pull request #2492 from Meliodas417/master

Update 0150.逆波兰表达式求值.md

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()
+
 
 ```