1 """ 2 Evaluate the value of an arithmetic expression in Reverse Polish Notation. 3 Valid operators are +, -, *, /. Each operand may be an integer or another expression. 4 Note: 5 Division between two integers should truncate toward zero. 6 The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation. 7 Example 1: 8 Input: ["2", "1", "+", "3", "*"] 9 Output: 9 10 Explanation: ((2 + 1) * 3) = 9 11 Example 2: 12 Input: ["4", "13", "5", "/", "+"] 13 Output: 6 14 Explanation: (4 + (13 / 5)) = 6 15 Example 3: 16 Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] 17 Output: 22 18 Explanation: 19 ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 20 = ((10 * (6 / (12 * -11))) + 17) + 5 21 = ((10 * (6 / -132)) + 17) + 5 22 = ((10 * 0) + 17) + 5 23 = (0 + 17) + 5 24 = 17 + 5 25 = 22 26 """ 27 """ 28 此题思路很简单，用栈即可实现，自己AC，但遇到三个麻烦 29 第一个是加减乘除操作要将字符型转为整型 30 第二个是被减数和减数的问题。先弹出的是减数，后弹出的是被减数。除法也需这样考虑 31 第三个是python向下取整的问题 -6 // 132 == -1 32 """ 33 class Solution1: 34 def evalRPN(self, tokens): 35 if not tokens: 36 return 0 37 stack = [] 38 for i in range(len(tokens)): 39 stack.append(tokens[i]) 40 if tokens[i] == '+': 41 stack.pop() 42 x = int(stack.pop()) 43 y = int(stack.pop()) 44 stack.append(y+x) 45 if tokens[i] == '-': 46 stack.pop() 47 x = int(stack.pop()) 48 y = int(stack.pop()) 49 stack.append(y-x) #减法注意先出栈的是减数,同除法 50 if tokens[i] == '*': 51 stack.pop() 52 x = int(stack.pop()) 53 y = int(stack.pop()) 54 stack.append(y*x) 55 if tokens[i] == '/': 56 stack.pop() 57 x = int(stack.pop()) 58 y = int(stack.pop()) 59 if x > 0 and y < 0 or x < 0 and y > 0: 60 stack.append(-(abs(y)//abs(x))) 61 else: #6//-132 == -1,python的向下取整,所以加了这条语句 62 stack.append(y//x) 63 return stack.pop() 64 65 if __name__ == '__main__': 66 ans = Solution1() 67 x = ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] 68 s = ans.evalRPN(x) 69 print(s) 70 71 print(6//-132)