Description
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? = 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
Constraints:
- 1 <= s.length <= 100
- s consists of digits 0-9 and characters ‘+’, ‘-’, ‘*’, ‘/’, ‘(’, and ‘)’.
- It is guaranteed that parentheses expression s is a VPS.
分析
题目的意思是：判断一个字符串里面括号的深度，思路也很直接，遍历的时候维护最大深度值depth就行了，t保存的是当前的左括号的最大值。我看了一下别人的实现，思路一样的。
代码
class Solution:def maxDepth(self, s: str) -> int:depth=0t=0for item in s:if(item=='('):t+=1depth=max(t,depth)elif(item==')'):t-=1depth=max(t,depth)return depth