Question

1021. Remove Outermost Parentheses

中文文档

Description

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 = P1 + P2 + ... + Pk, where Pi 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: s = "(()())(())"
Output: "()()()"
Explanation: 
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

Input: s = "(()())(())(()(()))"
Output: "()()()()(())"
Explanation: 
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Example 3:

Input: s = "()()"
Output: ""
Explanation: 
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

 

Constraints:

• 1 <= s.length <= 105
• s[i] is either '(' or ')'.
• s is a valid parentheses string.

Solutions

Solution 1

class Solution:
  def removeOuterParentheses(self, s: str) -> str:
    ans = []
    cnt = 0
    for c in s:
      if c == '(':
        cnt += 1
        if cnt > 1:
          ans.append(c)
      else:
        cnt -= 1
        if cnt > 0:
          ans.append(c)
    return ''.join(ans)

class Solution {
  public String removeOuterParentheses(String s) {
    StringBuilder ans = new StringBuilder();
    int cnt = 0;
    for (int i = 0; i < s.length(); ++i) {
      char c = s.charAt(i);
      if (c == '(') {
        if (++cnt > 1) {
          ans.append(c);
        }
      } else {
        if (--cnt > 0) {
          ans.append(c);
        }
      }
    }
    return ans.toString();
  }
}

class Solution {
public:
  string removeOuterParentheses(string s) {
    string ans;
    int cnt = 0;
    for (char& c : s) {
      if (c == '(') {
        if (++cnt > 1) {
          ans.push_back(c);
        }
      } else {
        if (--cnt) {
          ans.push_back(c);
        }
      }
    }
    return ans;
  }
};

func removeOuterParentheses(s string) string {
  ans := []rune{}
  cnt := 0
  for _, c := range s {
    if c == '(' {
      cnt++
      if cnt > 1 {
        ans = append(ans, c)
      }
    } else {
      cnt--
      if cnt > 0 {
        ans = append(ans, c)
      }
    }
  }
  return string(ans)
}

function removeOuterParentheses(s: string): string {
  let res = '';
  let depth = 0;
  for (const c of s) {
    if (c === '(') {
      depth++;
    }
    if (depth !== 1) {
      res += c;
    }
    if (c === ')') {
      depth--;
    }
  }
  return res;
}

impl Solution {
  pub fn remove_outer_parentheses(s: String) -> String {
    let mut res = String::new();
    let mut depth = 0;
    for c in s.chars() {
      if c == '(' {
        depth += 1;
      }
      if depth != 1 {
        res.push(c);
      }
      if c == ')' {
        depth -= 1;
      }
    }
    res
  }
}

Solution 2

class Solution:
  def removeOuterParentheses(self, s: str) -> str:
    ans = []
    cnt = 0
    for c in s:
      if c == '(':
        cnt += 1
      if cnt > 1:
        ans.append(c)
      if c == ')':
        cnt -= 1
    return ''.join(ans)

class Solution {
  public String removeOuterParentheses(String s) {
    StringBuilder ans = new StringBuilder();
    int cnt = 0;
    for (int i = 0; i < s.length(); ++i) {
      char c = s.charAt(i);
      if (c == '(') {
        ++cnt;
      }
      if (cnt > 1) {
        ans.append(c);
      }
      if (c == ')') {
        --cnt;
      }
    }
    return ans.toString();
  }
}

class Solution {
public:
  string removeOuterParentheses(string s) {
    string ans;
    int cnt = 0;
    for (char& c : s) {
      if (c == '(') {
        ++cnt;
      }
      if (cnt > 1) {
        ans.push_back(c);
      }
      if (c == ')') {
        --cnt;
      }
    }
    return ans;
  }
};

func removeOuterParentheses(s string) string {
  ans := []rune{}
  cnt := 0
  for _, c := range s {
    if c == '(' {
      cnt++
    }
    if cnt > 1 {
      ans = append(ans, c)
    }
    if c == ')' {
      cnt--
    }
  }
  return string(ans)
}