LeetCode 1111. Maximum Nesting Depth of Two Valid Parentheses Strings(java)

A string is a valid parentheses string (denoted VPS) if and only if it consists of “(” and “)” characters only, and:

It is the empty string, 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(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 seq, split it into two disjoint nonempty subsequences A and B, such that A and B are VPS’s (and A.length + B.length = seq.length).

Now choose any such A and B such that max(depth(A), depth(B)) is the minimum possible value.

Return an answer array (of length seq.length) that encodes such a choice of A and B: answer[i] = 0 if seq[i] is part of A, else answer[i] = 1. Note that even though multiple answers may exist, you may return any of them.

Example 1:

Input: seq = "(()())"
Output: [0,1,1,1,1,0]

Example 2:

Input: seq = "()(())()"
Output: [0,0,0,1,1,0,1,1]

Constraints:

1 <= seq.size <= 10000

思路：先找出左右括號相差最大的差值點，然後重新iterate，當遇到左右括號差值大於最大差值點的一半的“(”，選用1，否則都選用0.

class Solution {
    public int[] maxDepthAfterSplit(String seq) {
        int len = seq.length();
        int max = 0;
        int num = 0;
        for (int i = 0; i < len; i++) {
            if (seq.charAt(i) == '(') {
                num++;
            } else {
                num--;
            }
            max = Math.max(max, Math.abs(num));
        }
        int pair = (max + 1) / 2;
        int l = 0;
        int r = 0;
        int[] res = new int[len];
        for (int i = 0; i < len; i++) {
            if (seq.charAt(i) == '(') {
                if (l < pair) {
                    res[i] = 0;
                } else {
                    res[i] = 1;
                    r++;
                }
                l++;
            } else {
                if (r > 0) {
                    res[i] = 1;
                    r--;
                } else res[i] = 0;
            }
        }
        return res;
    }
}