描述
Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules:
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. An empty string is also valid.
Example 1:
Input: "()"
Output: True
Example 2:
Input: "(*)"
Output: True
Example 3:
Input: "(*))"
Output: True
Note:
The string size will be in the range [1, 100].
解析
根据题意，主要判断字符串是否合法，合法的依据就是左右圆括号能够前后配对，而难点在于 * 可以当作 ( 或者 ) 存在，传统的栈方法不太适合，而观看大神们的解答很巧妙：
遍历 s 中的所有字符 c，利用集合 set 记录当前左括号-右括号的差值
若 c == '('，将 vset 替换为其所有元素 +1
若 c == ')'，将 vset 替换为其所有不小于 1 的元素 - 1 ，这里巧妙地解决了类似 ")(" 的情况
若 c == '*’，将 vset 中的所有元素 +1 ，-1 后，保留非负元素
最后看是否在集合中包含 0 ，有 0 则说明字符串中的左右括号和能够替换括号的 * 都能够一一合法配对的情况是存在的
解答
class Solution(object):
def checkValidString(s):
old = set([0])
for c in s:
new = set()
if c=="(":
for i in old:
new.add(i+1)
elif c==")":
for i in old:
if i>0:
new.add(i-1)
elif c=="*":
for i in old:
new.add(i+1)
new.add(i)
if i>0:
new.add(i-1)
old = new
return 0 in old
运行结果
Runtime: 16 ms, faster than 85.10% of Python online submissions for Valid Parenthesis String.
Memory Usage: 12.8 MB, less than 34.38% of Python online submissions for Valid Parenthesis String.
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