LeetCode 150题，逆波兰表达式求值，用栈解决，怎么算？

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评估逆波兰表达式（RPN）的值。有效的运算符包括：+、-、*、/。每个操作数可以是整数或另一个表达式。注意：两个整数之间的除法应截断到零。给定的R...

Evaluate the value of an arithmetic expression inReverse Polish Notation.

Valid operators are+,-,*,/. Each operand may be an integer or another expression.

Note:

• Division between two integers should truncate toward zero.
• The given RPN expression is always valid. That means the expression would always evaluate to a result and there won‘tbe anydivideby zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"] Output: 9 Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"] Output: 6 Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] Output: 22 Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 = ((10 * (6 / (12 * -11))) + 17) + 5 = ((10 * (6 / -132)) + 17) + 5 = ((10 * 0) + 17) + 5 = (0 + 17) + 5 = 17 + 5 = 22

这个题目就是用stack，如果看到{‘+‘, ‘-‘,‘*‘,‘/‘}, 将stack pop两次，然后进行相应的计算，注意 1// -32 == -1， 而 int(1/ -32) == 0;

T: O(n) S: O(n)

Code

class Solution: def evalRPN(self, tokens: List[str]) -> int: if not tokens: return 0 stack, d = [], {‘+‘, ‘-‘,‘*‘,‘/‘} for token in tokens: if token in d: num2 = stack.pop() num1 = stack.pop() if token == ‘+‘: num = num1 + num2 elif token == ‘-‘: num = num1 - num2 elif token == ‘*‘: num = num1 * num2 else: num = int(num1/num2) stack.append(num) else: stack.append(int(token)) return stack[0]