LeetCode 1614. Maximum Nesting Depth of the Parentheses

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

Constraints:

• 1 <= s.length <= 100
• s consists of digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.
• It is guaranteed that parentheses expression s is a VPS.

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这题是要求一个表达式里最深的nested括号有多深，嗯，还是很典型stack的思想。刚开始想了一下到底有没有必要用到stack，但还是先用了一下stack。其实就只需要遇到(就往里push遇到)就pop，每次push完Math.max()判断一下是不是最深就行。

class Solution {
    public int maxDepth(String s) {
        int result = 0;
        Deque stack = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c == '(') {
                stack.push(c);
                result = Math.max(result, stack.size());
            } else if (c == ')') {
                stack.pop();
            }
        }
        return result;
    }
}

又仔细想了一下其实直接用个int存就行了，因为只care它的size。瞬间时间空间效率都上去了。

class Solution {
    public int maxDepth(String s) {
        int result = 0;
        int curr = 0;
        for (char c : s.toCharArray()) {
            if (c == '(') {
                curr++;
                result = Math.max(result, curr);
            } else if (c == ')') {
                curr--;
            }
        }
        return result;
    }
}