LeetCode题解(0150):逆波兰表达式求值(Python)

题目:原题链接(中等)

标签:栈

解法 时间复杂度 空间复杂度 执行用时
Ans 1 (Python) O ( N ) O(N) O ( N ) O(N) 44ms (89.77%)
Ans 2 (Python)
Ans 3 (Python)

LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。

解法一:

def evalRPN(self, tokens: List[str]) -> int:
    stack = []
    for token in tokens:
        if token == "+":
            b = stack.pop()
            a = stack.pop()
            stack.append(a + b)
        elif token == "-":
            b = stack.pop()
            a = stack.pop()
            stack.append(a - b)
        elif token == "*":
            b = stack.pop()
            a = stack.pop()
            stack.append(a * b)
        elif token == "/":
            b = stack.pop()
            a = stack.pop()
            stack.append(int(a / b))
        else:
            stack.append(int(token))

    return stack[0]

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转载自blog.csdn.net/Changxing_J/article/details/107756307