Nesting depth LeetCode 1111. effective parentheses (separated parity)

1. Topic

Bracket only by valid string "("and ")"configured, and in accordance with one of several of the following conditions:

• Null string
• Connection may be referred to as AB (A and B are connected), wherein A and B are valid string brackets
• Nesting, can be written as (A), where A is a valid string brackets

Similarly, we can define the effective depth of nesting brackets string s depth (S):

• s is empty,depth("") = 0
• When s is connected to A and B, depth(A + B) = max(depth(A), depth(B))where A and B are valid string brackets
• s is nested case depth("(" + A + ")") = 1 + depth(A)where A is a valid string brackets
  example: ""，"()()"，和 "()(()())"are valid string brackets, respectively, 0,1,2 nesting depth, and ")(" 和 "(()"is not effective in parentheses string.

To give you a valid string bracket seq, dividing it into two disjoint sub-sequences A and B, and A and B satisfy the brackets define a valid character string (note: A.length + B.length = seq.length).

Now, you need to choose the arbitrary set of valid strings brackets A and B, make max(depth(A), depth(B))possible the value of the minimum .

Returns the length of seq.lengththe answer array answer, select A or B encoding rule is: if seq [i] is the A part, then the answer [i] = 0. Otherwise, answer [i] = 1.
Even if there are multiple answers exist to meet the requirements, you also need a return.

示例 1：
输入：seq = "(()())"
输出：[0,1,1,1,1,0]

示例 2：
输入：seq = "()(())()"
输出：[0,0,0,1,1,0,1,1]
 
提示：
1 <= text.size <= 10000

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

2. Problem Solving

• Title mean, as much as possible so that a minimum of two open brackets maximum nesting depth
• Then according to the depth into odd layers, even layers, respectively, out on the line

class Solution {
public:
    vector<int> maxDepthAfterSplit(string seq) {
    	int i, j = 0, depth = 0;
    	vector<int> ans(seq.size(),0);
        char prev = '*';
    	for(i = 0; i < seq.size(); ++i,++j)
    	{
            if(prev == ')')
                depth--;
    		if(seq[i] == '(')
    			depth++;
    		if(depth & 1)//奇数层
                ans[j] = 1;
            prev = seq[i];
            // cout << depth << " ";
    	}
    	return ans;
    }
};