0150.逆波兰表达式求值 添加更快速的python解

This commit is contained in:
Frank Mao
2023-01-06 00:35:56 +08:00
parent 2863a699cc
commit abedcbbe04

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@ -163,6 +163,25 @@ class Solution {
python3
```python
from operator import add, sub, mul
class Solution:
op_map = {'+': add, '-': sub, '*': mul, '/': lambda x, y: int(x / y)}
def evalRPN(self, tokens: List[str]) -> int:
stack = []
for token in tokens:
if token not in {'+', '-', '*', '/'}:
stack.append(int(token))
else:
op2 = stack.pop()
op1 = stack.pop()
stack.append(self.op_map[token](op1, op2)) # 第一个出来的在运算符后面
return stack.pop()
```
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: