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 "" + "" = "".解题要点:
count左右括号出现次数,如左,count加一,如右,count减少一。如count等于0时,先取出这段有效括号,再处理去掉外层括号,加进StringBuffer,最后更新下一次的起点。返回sb.toString()。
class Solution {
public String removeOuterParentheses(String S) {
int count = 0, start = 0;
StringBuffer sb = new StringBuffer();
for(int i = 0; i < S.length(); i++){
if(S.charAt(i)== '(') count++;
else if(S.charAt(i) == ')') count--;
if(count == 0){
String temp = S.substring(start, i + 1);
temp = temp.substring(1, temp.length() - 1);
sb.append(temp);
start = i + 1;
}
}
return sb.toString();
}
}Last updated
Was this helpful?