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(Aconcatenated withB), whereAandBare VPS’s, or - It can be written as
(A), whereAis a VPS.
We can similarly define the nesting depth depth(S) of any VPS S as follows:
depth("") = 0depth(A + B) = max(depth(A), depth(B)), whereAandBare VPS’sdepth("(" + A + ")") = 1 + depth(A), whereAis 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.
Example 1:
Input: seq = "(()())"
Output: [0,1,1,1,1,0]
Example 2:
Input: seq = "()(())()"
Output: [0,0,0,1,1,0,1,1]
这道题给了一个合法的括号字符串，定义了一种括号深度，就是最深的括号嵌套层数。现在让将这个括号字符串拆分成为两个合法的括号字符串，且二者之中的较大深度最小，就是说要尽可能让二者的深度相同。返回数组中，用0和1来区分不同的字符串。既然是尽可能的平均拆分，那么拆分后的每个字符串的深度一定不会超过原来的一半，就是说有嵌套括号时就要平均分配给两个字符串。什么时候会有嵌套括号呢，就比如 “(())” 这种，要拆分为 “()” 和 “()”，而对于没有嵌套括号的，比如 “()()()” 这种，可以都放到一个字符串中都没问题。所以问题的关键就是对于连续的左括号，要将其平均的分配到不同的字符串中。所以处理的方法就是使用一个 level 变量，初始化为0，然后遍历给定括号字符串，若遇到了左括号，则 level 对2取余，将结果存入 res 中，为了避免连续左括号加入同一个组，将 level 自增1。这样接下来又遇到左括号时，就可以加进不同的组，若接下来遇到右括号了，则应该先给 level 自增1，再对2取余，这样就可以跟前一个左括号划分到同一个组中了，参见代码如下：
解法一：
class Solution {
public:
vector<int> maxDepthAfterSplit(string seq) {
int n = seq.size(), level = 0;
vector<int> res(n);
for (int i = 0; i < n; ++i) {
if (seq[i] == '(') {
res[i] = level++ % 2;
} else {
res[i] = ++level % 2;
}
}
return res;
}
};
下面这种写法更加的简洁，但其实思路都是一样的，只不过用i代替了上面的 level 变量的作用了，参见代码如下：
解法二：
class Solution {
public:
vector<int> maxDepthAfterSplit(string seq) {
vector<int> res(seq.size());
for (int i = 0; i < seq.size(); ++i) {
res[i] = seq[i] == '(' ? i & 1 : (1 - i & 1);
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1111
类似题目：
Maximum Nesting Depth of the Parentheses
参考资料：
https://leetcode.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/
LeetCode All in One 题目讲解汇总(持续更新中…)
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