Easy
A string is a valid parentheses string (denoted VPS) if it meets one of the following:
""
, or a single character not equal to "("
or ")"
,AB
(A
concatenated with B
), where A
and B
are VPS’s, or(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
Constraints:
1 <= s.length <= 100
s
consists of digits 0-9
and characters '+'
, '-'
, '*'
, '/'
, '('
, and ')'
.s
is a VPS.import java.util.ArrayDeque;
import java.util.Deque;
public class Solution {
public int maxDepth(String s) {
Deque<Character> stack = new ArrayDeque<>();
int maxDepth = 0;
for (char c : s.toCharArray()) {
if (c == '(') {
stack.push(c);
} else if (c == ')') {
maxDepth = Math.max(maxDepth, stack.size());
stack.pop();
}
}
return maxDepth;
}
}