22. Generate Parentheses

Generate Parentheses - LeetCode

 

1. Brute Force (특정 길이의 괄호 문자열을 전부 만들고 적합한지 판단)

from typing import *


class Solution:
    answer = []

    def generateParenthesis(self, n: int) -> List[str]:
        self.answer=[]
        self.dfs("", n)
        return self.answer
    def dfs(self, cur, target):
        if len(cur) == target * 2:
            if self.check(cur):
                self.answer.append(cur)
            return
        else:
            self.dfs(cur + "(", target)
            self.dfs(cur + ")", target)

    def check(self, input_str):
        stack = []
        for i in range(len(input_str)):
            if not stack or input_str[i] == "(":
                stack.append(input_str[i])
            else:
                if not stack or stack[-1] != "(":
                    return False
                else:
                    stack.pop()
        if not stack:
            return True
        return False

2. 창의적인 Backtracking

from typing import *


class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        answer = []

        def dfs(string_list, left, right):
            if len(string_list) == 2 * n:
                answer.append("".join(string_list))
                return

            if left < n:
                string_list.append("(")
                dfs(string_list, left + 1, right)
                string_list.pop()
            if right < left:
                string_list.append(")")
                dfs(string_list, left, right + 1)
                string_list.pop()

        dfs([], 0, 0)
        return answer