谁能帮我改写LeetCode中等难度题150.逆波兰表达式求值？

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pythonclass Solution(object): def evalRPN(self, tokens): stack=deque() for token in tokens: if token=='+': num2=stack.pop() num1=stack.pop() stack.append(num1 + num2) elif token=='-': num2=stack.pop() num1=stack.pop() stack.append(num1 - num2) elif token=='*': num2=stack.pop() num1=stack.pop() stack.append(num1 * num2) elif token=='/': num2=stack.pop() num1=stack.pop() stack.append(int(num1 / num2)) else: stack.append(int(token)) return stack.pop()

mycode 42.30%、

注意：如果不考虑符号，-1//3=-1而不是等于0，因为是向下取整

class Solution(object): def evalRPN(self, tokens): """ :type tokens: List[str] :rtype: int """ from collections import deque def cal(data_1,data_2,item): if item == ‘/‘: return abs(data_2) // abs(data_1) *(-1 if (data_2 > 0) ^ (data_1 > 0) else 1) elif item == ‘+‘: return data_2 + data_1 elif item == ‘-‘: return data_2 - data_1 else: return data_2 * data_1 dq = deque() calset = [‘/‘,‘+‘,‘-‘,‘*‘] for item in tokens: if item in calset: #print(‘if‘) data_1 = dq.pop() data_2 = dq.pop() data = cal(int(data_1),int(data_2),item) dq.append(data) else: #print(‘else‘) dq.append(item) #print(dq) return dq[0]

参考：

#www.cnblogs.com/zuoyuan/p/3760530.html 注意负数的除法，c++和pytho的区别 class Solution: # @param tokens, a list of string # @return an integer def evalRPN(self, tokens): stack = [] for i in range(0,len(tokens)): if tokens[i] != ‘+‘ and tokens[i] != ‘-‘ and tokens[i] != ‘*‘ and tokens[i] != ‘/‘: stack.append(int(tokens[i])) else: a = stack.pop() b = stack.pop() if tokens[i] == ‘+‘: stack.append(a+b) if tokens[i] == ‘-‘: stack.append(b-a) if tokens[i] == ‘*‘: stack.append(a*b) if tokens[i] == ‘/‘: if a*b < 0: stack.append(-((-b)/a)) else: stack.append(b/a) return stack.pop()