[LeetCode] 1021. Remove Outermost Parentheses
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
题目大意是原字符串是有括号组成的,有单个括号,也可能有嵌套括号。将原字符串中的所有最外层括号去掉。比如"(())"会变成"()","(())(())"会变成"()()","()"会变成""。
因为要去掉最外层括号,所以将除最外层以外的"(",以及其配对的")"放入结果字符串中。使用flag标记")"是否是最外层括号的后半部分。
方法一:
我一开始想到了使用堆栈来存储"(",碰到")"时就将除最外层以外的括号都弹出到结果字符串中,然后使用一个flag标记遇到的")"是否是最外层括号的后半部分,如果是,就直接弹出而不放入结果字符串中。
代码如下:
class Solution { public: string removeOuterParentheses(string S) { string res = ""; stack<char> s; int flag = 0; for (int i = 0; i < S.size(); ++i) { if (S[i] == '(') { s.push(S[i]); } else if (S[i] == ')') { if (s.size() <= 1 && flag==0) { s.pop(); continue; } while (s.size() > 1) { res += s.top(); s.pop(); flag++; } res += S[i]; flag--; } } return res; } };
方法二:
但是使用flag的话其实根本没必要先将"("存入堆栈当中,只要不将最外层括号放入结果字符串中,遇到最外层")"时返回循环就好了。
代码如下:
class Solution { public: string removeOuterParentheses(string S) { string res = ""; int flag = 0; for (int i = 0; i < S.size(); ++i) { if (flag==0) { flag++; continue; } else if (S[i] == '(') { res += S[i]; flag++; } else if (S[i] == ')') { if (flag > 1) { res += S[i]; } flag--; } } return res; } };