【leetcode】1106. Parsing A Boolean Expression
作者:互联网
题目如下:
Return the result of evaluating a given boolean
expression
, represented as a string.An expression can either be:
"t"
, evaluating toTrue
;"f"
, evaluating toFalse
;"!(expr)"
, evaluating to the logical NOT of the inner expressionexpr
;"&(expr1,expr2,...)"
, evaluating to the logical AND of 2 or more inner expressionsexpr1, expr2, ...
;"|(expr1,expr2,...)"
, evaluating to the logical OR of 2 or more inner expressionsexpr1, expr2, ...
Example 1:
Input: expression = "!(f)" Output: trueExample 2:
Input: expression = "|(f,t)" Output: trueExample 3:
Input: expression = "&(t,f)" Output: falseExample 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'
标签:Expression,list,ts,1106,Boolean,evaluating,expr1,expression,stack 来源: https://www.cnblogs.com/seyjs/p/11139786.html