描述
A string is a valid parentheses string (denoted VPS) if it meets one of the following:
- It is an empty string "", or a single character not equal to "(" or ")",
- It can be written as AB (A concatenated with B), where A and B are VPS's, or
- It can be written as (A), where A is a VPS.
We can similarly define the nesting depth depth(S) of any VPS S as follows:
- depth("") = 0
- depth(C) = 0, where C is a string with a single character not equal to "(" or ")".
- depth(A + B) = max(depth(A), depth(B)), where A and B are VPS's.
- depth("(" + A + ")") = 1 + depth(A), where A is a VPS.
For example, "", "()()", and "()(()())" are VPS's (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS's.
Given a VPS represented as string s, return the nesting depth of s.
Example 1:
Input: s = "(1+(2*3)+((8)/4))+1"
Output: 3
Explanation: Digit 8 is inside of 3 nested parentheses in the string.
Example 2:
Input: s = "(1)+((2))+(((3)))"
Output: 3
Example 3:
Input: s = "1+(2*3)/(2-1)"
Output: 1
Example 4:
Input: s = "1"
Output: 0
Note:
- 1 <= s.length <= 100
- s consists of digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.
- It is guaranteed that parentheses expression s is a VPS.
解析
根据题意，遍历字符串的时候，只需要用栈来记录左括号，在遇到右括号的时候将栈顶的左括号去掉即可，同时记录栈中存在左括号的最大数量即可。
解答
class Solution(object):
def maxDepth(self, s):
"""
:type s: str
:rtype: int
"""
stack = []
res = 0
for c in s:
if c=='(':
stack.append(c)
res = max(res, len(stack))
elif c==')':
stack.pop()
return res
运行结果
Runtime: 20 ms, faster than 57.38% of Python online submissions for Maximum Nesting Depth of the Parentheses.
Memory Usage: 13.6 MB, less than 29.22% of Python online submissions for Maximum Nesting Depth of the Parentheses.
原题链接：leetcode.com/problems/ma…
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