# 算法练习Stack--P1021-easy

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 "” + "” = “”.

Runtime: 3 ms, faster than 69.58% of Java online submissions for Remove Outermost Parentheses. Memory Usage: 37 MB, less than 94.81% of Java online submissions for Remove Outermost Parentheses.

public String removeOuterParentheses(String S) {

StringBuilder sb = new StringBuilder();
int open=0, close=0, start=0;
for(int i=0; i<S.length(); i++) {
if(S.charAt(i) == '(') {
open++;
} else if(S.charAt(i) == ')') {
close++;
}
if(open==close) {
sb.append(S.substring(start+1, i));
start=i+1;
}
}
return sb.toString();
}