LeetCode #1021 RemoveOutermostParentheses. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
It takes just 20 min? but one of the worst performance.
Runtime: 48 ms, faster than 5.15% of C++ online submissions for Remove Outermost Parentheses.
Memory Usage: 8.9 MB, less than 90.00% of C++ online submissions for Remove Outermost Parentheses.
LeetCode #1021
Q.
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.
뭐라는지 잘 모르겠고, 그냥 각 괄호들 중에서 최외곽 괄호를 없애라. 예제 보면 바로 앎
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
Process
// Process
//1. Input string S
//2. Iterate from begin to the end
// 2.1. Check if it's (, or )
// 2.1.1. If it's ( and start one -> delete it
// 2.1.2. If it's ) and end one -> delete it
//3. Return result
// 처리과정
//1. 스트링 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. (, ) 어떤 것인지 확인해서
// 2.1.1. ( 이면서 시작하는 괄호이면 지운다.
// 2.1.2. ) 이면서 끝나는 괄호이면 지운다.
//3. 결과 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
string removeOuterParentheses(string S) {
int parenthCount = 0;
if (S.length() > 0) {
for (int i = 0; i < S.length(); ++i) {
if (S[i] == '(') {
if (parenthCount++ == 0) {
S.erase(S.begin() + (i--));
}
}
else {
if (--parenthCount == 0) {
S.erase(S.begin() + (i--));
}
}
}
}
return S;
}
};
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