[leetcode] 1614. Maximum Nesting Depth of the Parentheses

[leetcode] 1614. Maximum Nesting Depth of the Parentheses

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, orIt 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("") = 0depth© = 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: 3Explanation: 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 <= 100s 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=0 t=0 for item in s: if(item=='('): t+=1 depth=max(t,depth) elif(item==')'): t-=1 depth=max(t,depth) return depth

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