（python) leetcode刷题——evaluate-reverse-polish-notation

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题目：
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are+,-,,/. Each operand may be an integer or another expression.
Some examples:
[“2”, “1”, “+”, “3”, ""] -> ((2 + 1) * 3) -> 9
[“4”, “13”, “5”, “/”, “+”] -> (4 + (13 / 5)) -> 6

解析：
典型的使用栈的例子，本科数据结构课程有讲。
1，遍历列表，建立空栈
2，遇到数字，将它入栈
3，遇到运算符，取栈内数字两次，得到的两个数进行对应运算符运算，将运算结果入栈
4，遍历完毕，取栈，得到最终结果

代码：

def prnval(a):
    stack = []
    for i in range(len(a)):
        if a[i] == '+' or a[i] == '-' or a[i] == '*' or a[i] == '/':
            x = stack.pop()
            y = stack.pop()
            if a[i] == '+':
                stack.append(x+y)
            elif a[i] == '-':
                stack.append(y-x)
            elif a[i] == '*':
                stack.append(x*y)
            else:
                stack.append(y//x)
        elif a[i] != None:
            stack.append(int(a[i]))
    return stack.pop()

全部代码：

def prnval(a):
    stack = []
    for i in range(len(a)):
        if a[i] == '+' or a[i] == '-' or a[i] == '*' or a[i] == '/':
            x = stack.pop()
            y = stack.pop()
            if a[i] == '+':
                stack.append(x+y)
            elif a[i] == '-':
                stack.append(y-x)
            elif a[i] == '*':
                stack.append(x*y)
            else:
                stack.append(y//x)
        elif a[i] != None:
            stack.append(int(a[i]))
    return stack.pop()

if __name__=='__main__':
    a = ["2","1","+","3","*"]
    b = ["4", "13", "5", "/", "+"]
    print(prnval(a),prnval(b)