Dec 27, 2024 · Nov 14, 2024 · Nov 16, 2024 · Dec 26, 2024 · Dec 26, 2024 · Dec 27, 2024
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
Original file line number	Diff line number	Diff line change
Expand Up		@@ -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:
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