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:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
思路:
1.暴力法
2.DFS回溯法
2的代码:
class Solution(object):
def generateParenthesis(self, n):
"""
:type n: int
:rtype: List[str]
"""
self.res = []
self._gen(0, 0, n, "")
return self.res
def _gen(self,left, right, n, cr):
if left == n and right == n:
self.res.append(cr)
return
if left < n:
self._gen(left + 1, right, n, cr + "(")
if left > right and right < n:
self._gen(left, right + 1, n, cr + ")")
大佬写法:(带DP的递归)
class Solution:
def __init__(self):
self.dp = dict(zip((0, 1), ([''], ['()'])))
def generateParenthesis(self, n):
"""
:type n: int
:rtype: List[str]
"""
if n not in self.dp:
self.dp[n] = ['(' + sl + ')' + sr for m in range(n) for sl in self.generateParenthesis(m) for sr in self.generateParenthesis((n - 1) - m)]
return self.dp[n]