Problem
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 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.
Example1
Input: seq = “(()())”
Output: [0,1,1,1,1,0]
Example2
Input: seq = “()(())()”
Output: [0,0,0,1,1,0,1,1]
Solution
字都认得，就是不知道题啥意思。看了解析才会。
思路：
- 根据 depth(A + B) = max(depth(A), depth(B)) 这个定义，整体的「嵌套深度」取决于子序列的「嵌套深度」的最大者；
- 要使得 max(depth(A), depth(B)) 的可能取值最小，分析示例的时候提到这很像一棵二叉树，要使得二叉树的深度最小，那么就需要该二叉树平衡，一个可行的做法是：把栈中连续出现的左括号 ( 根据奇偶性分到不同的组，右括号随与之匹配左括号的组号；
- 如果出现 () 这种子序列，即左括号后面连着出现了右括号，其实分在那一组都是没有关系的，因为它的存在不会使得「嵌套深度」更深。
把深度为1和3的分给A，深度为2的分给B。
class Solution {
public:
vector<int> maxDepthAfterSplit(string seq) {
int d = 0;
vector<int> ans;
for (char& c : seq)
if (c == '(') {
++d;
ans.push_back(d % 2);
}
else {
ans.push_back(d % 2);
--d;//注意
}
return ans;
}
};
Ref
https://leetcode-cn.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/solution/qian-tao-shen-du-wan-cheng-gua-hao-pi-pei-wen-ti-s/
