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  • 【leetcode】1106. Parsing A Boolean Expression

    题目如下:

    Return the result of evaluating a given boolean expression, represented as a string.

    An expression can either be:

    • "t", evaluating to True;
    • "f", evaluating to False;
    • "!(expr)", evaluating to the logical NOT of the inner expression expr;
    • "&(expr1,expr2,...)", evaluating to the logical AND of 2 or more inner expressions expr1, expr2, ...;
    • "|(expr1,expr2,...)", evaluating to the logical OR of 2 or more inner expressions expr1, expr2, ...

    Example 1:

    Input: expression = "!(f)"
    Output: true
    

    Example 2:

    Input: expression = "|(f,t)"
    Output: true
    

    Example 3:

    Input: expression = "&(t,f)"
    Output: false
    

    Example 4:

    Input: expression = "|(&(t,f,t),!(t))"
    Output: false
    

    Constraints:

    • 1 <= expression.length <= 20000
    • expression[i] consists of characters in {'(', ')', '&', '|', '!', 't', 'f', ','}.
    • expression is a valid expression representing a boolean, as given in the description.

    解题思路:本题和表达式运算的题目相似。遍历expression并将每一个字符依次入栈,如果遇到')',则找出离栈顶最近的'(',计算出括号之内的表达式的值并将该值入栈,直到expression遍历完成为止。

    代码如下:

    class Solution(object):
        def parseBoolExpr(self, expression):
            """
            :type expression: str
            :rtype: bool
            """
            stack = []
            expression = expression.replace('t','1')
            expression = expression.replace('f', '0')
            ex = list(expression)
            while len(ex) > 0:
                char = ex.pop(0)
                if char != ')':
                    stack.append(char)
                    continue
                ts = ''
                while len(stack) > 0:
                    item = stack.pop(-1)
                    if item == '(':
                        break
                    ts += item
                ts_list = ts.split(',')
                and_or = stack.pop(-1)
                if and_or == '!':
                    stack.append('1' if ts_list[0] == '0' else '0' )
                elif and_or == '|':
                    stack.append('1' if '1' in ts_list else '0')
                else:
                    stack.append('0' if '0' in ts_list else '1')
            return stack[0] == '1'
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  • 原文地址:https://www.cnblogs.com/seyjs/p/11139786.html
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