A valid parentheses string is either empty ("")
, "(" + A + ")"
, or A + B
, where A
and B
are valid parentheses strings, and +
represents string concatenation. For example, ""
, "()"
, "(())()"
, and "(()(()))"
are all valid parentheses strings.
A valid parentheses string S
is primitive if it is nonempty, and there does not exist a way to split it into S = A+B
, with A
and B
nonempty valid parentheses strings.
Given a valid parentheses string S
, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k
, where P_i
are primitive valid parentheses strings.
Return S
after removing the outermost parentheses of every primitive string in the primitive decomposition of S
.
Example 1:
Input: "(()())(())" Output: "()()()" Explanation: The input string is "(()())(())", with primitive decomposition "(()())" + "(())". After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input: "(()())(())(()(()))" Output: "()()()()(())" Explanation: The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))". After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input: "()()" Output: "" Explanation: The input string is "()()", with primitive decomposition "()" + "()". After removing outer parentheses of each part, this is "" + "" = "".
Note:
S.length <= 10000
S[i]
is"("
or")"
S
is a valid parentheses string
Companies:
Google
Related Topics:
Stack
// OJ: https://leetcode.com/problems/remove-outermost-parentheses/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
string removeOuterParentheses(string S) {
string ans;
int cnt = 0, start = 0;
for (int i = 0; i < S.size(); ++i) {
if (S[i] == '(') ++cnt;
else if (--cnt == 0) {
ans += S.substr(start + 1, i - start - 1);
start = i + 1;
}
}
return ans;
}
};