给出 n 代表生成括号的对数,请你写出一个函数,使其能够生成所有可能的并且有效的括号组合。
例如,给出 n = 3,生成结果为:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
class Solution:
def generateParenthesis(self, n):
"""
:type n: int
:rtype: List[str]
"""
self.list = []
self.recursion(0, 0, n, "")
return self.list
def recursion(self,left, right, n, result):
if left== n and right == n:
self.list.append(result)
return
if left<n:
self.recursion(left+1,right,n,result+"(")
if left>right and right<left:
self.recursion(left,right+1,n,result+")")