BGM（？）：圣堂之门-阿沁《梵谷的左耳》
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
Constraints:
1 <= s.length <= 100
s consists of digits 0-9 and characters ‘+’, ‘-’, ‘*’, ‘/’, ‘(’, and ‘)’.
It is guaranteed that parentheses expression s is a VPS.
Solution
Try to think it easy: define s for the number of (, and deep for the deep
- When visit’(', just count start.
- When visit’)', regarding as a VPS finish, compare current deep with start. Finally, sub the start as the ‘layer’ of this nest has been finish
Code
class Solution {
public:int maxDepth(string s){if(s==" "||(s.length()==1&&s!="("&&s!=")")) return 0;else{int deep=0,start=0;for(int i=0;i<s.size();++i){if(s[i]=='('){++start;}else if(s[i]==')') {deep=max(deep,start);--start;}}return deep;}return 0;}
};
