[leetcode] 150. Evaluate Reverse Polish Notation @ python

最新推荐文章于 2021-03-20 22:13:41 发布

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本文介绍了一种使用栈来解析并计算逆波兰表达式的算法。逆波兰表达式是一种后缀表达式，常见于计算器中，可以避免使用括号。文章详细解释了如何通过遍历表达式中的每个元素，将其压入栈中或从栈中取出两个元素进行运算，最终得到表达式的值。

原题

Evaluate the value of an arithmetic expression in Reverse 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’t be any divide by 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

解法

stack. 构造nums列表存放数字, 遍历tokens, 如果是数字, 那么把它放到nums里, 如果是运算符号, 那么提取最后两个数字进行运算后, 再放进nums. 注意题目要求整除相除时向0取整, 那么当整除结果为负且无法除尽时, 例如:在python里, -4//3 = -2, 此时我们需要加1得出符合题意的结果.
Time: O(n)
Space: O(n)

代码

class Solution(object):
    def evalRPN(self, tokens):
        """
        :type tokens: List[str]
        :rtype: int
        """
        nums = []
        for token in tokens:
            if token.isdigit() or token.lstrip('-').isdigit():
                nums.append(int(token))
            elif token == '+':
                ans = nums.pop() + nums.pop()
                nums.append(ans)
            elif token == '-':
                last = nums.pop()
                first = nums.pop()
                ans = first - last
                nums.append(ans)
            elif token == '*':
                ans = nums.pop() * nums.pop()
                nums.append(ans)
            elif token == '/':
                last = nums.pop()
                first = nums.pop()
                ans = first // last
                if ans < 0 and first % last != 0:
                    ans += 1
                nums.append(ans)
        return nums[0]