動機

Problem

Given a string s containing only three types of characters: '(', ')' and '*', return true if s is valid.

The following rules define a valid string:

  • Any left parenthesis '(' must have a corresponding right parenthesis ')'.
  • Any right parenthesis ')' must have a corresponding left parenthesis '('.
  • Left parenthesis '(' must go before the corresponding right parenthesis ')'.
  • '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string .

 

Example 1:

Input: s = ()Output: true

Example 2:

Input: s = (*)Output: true

Example 3:

Input: s = (*))Output: true

 

Constraints:

  • 1 <= s.length <= 100
  • s[i] is '(', ')' or '*'.

Sol

重點是怎麼處理星號，

  • 左括號只能和他之後的星號配對
  • 右括號只能和他之前的星號配對
class Solution:
    def checkValidString(self, s: str) -> bool:
        left = []
        star = []
        right = []
        for i in range(len(s)):
            if s[i] == '*':
                star.append(i)
            elif s[i] == '(':
                left.append(i)
            else:
                if left:
                    left.pop()
                elif star:
                    star.pop()
                else:
                    right.append(i)
        print(left,star,right)
        if left:
            while left and star and left[-1] < star[-1]:
                left.pop()
                star.pop()
        elif right:
            while right and star and star[-1] < right[-1]:
                right.pop()
                star.pop()
            
        print(left,star,right)
        return not left and not right

case study

  1. 把所有星號當成左括號，去計數，如果左括號太少，就直接False
  2. 把所有星號當成右括號，去計數，全部跑完，看有沒有被扣完 (多扣沒關係，因為在第一條已經證明左括號+星號一定足夠匹配真正的右括號)
class Solution:
    def checkValidString(self, s: str) -> bool:
        cmin = cmax = 0
        for c in s:
            cmax = cmax - 1 if c == ')' else cmax + 1
            cmin = cmin + 1 if c == '(' else max(cmin - 1, 0)
            if cmax < 0:
                return False
        return cmin == 0