1021. Remove Outermost Parentheses

# 1021. Remove Outermost Parentheses#

## 题目 #

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

## 代码 #

``````
package leetcode

// 解法一
func removeOuterParentheses(S string) string {
now, current, ans := 0, "", ""
for _, char := range S {
if string(char) == "(" {
now++
} else if string(char) == ")" {
now--
}
current += string(char)
if now == 0 {
ans += current[1 : len(current)-1]
current = ""
}
}
return ans
}

// 解法二
func removeOuterParentheses1(S string) string {
stack, res, counter := []byte{}, "", 0
for i := 0; i < len(S); i++ {
if counter == 0 && len(stack) == 1 && S[i] == ')' {
stack = stack[1:]
continue
}
if len(stack) == 0 && S[i] == '(' {
stack = append(stack, S[i])
continue
}
if len(stack) > 0 {
switch S[i] {
case '(':
{
counter++
res += "("
}
case ')':
{
counter--
res += ")"
}
}
}
}
return res
}

``````

Apr 8, 2023