Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
产生配对的括号匹配序列
给定n,则最终输出的括号序列为2*n,假定最终序列是由括号依次放入,则假定左括号个数为opennum,右括号个数为closenum,当依次放入括号:
初始时,应该放入左括号 或者当opennum==closenum时,应该放入左括号
当opennum<n时,均可以放入左括号
当opennum>closenum时,说明此时左括号多于右括号,即此时应该放入右括号
当opennum<closenum时,此时已经不匹配,则停止放入
初始时,首先放入左括号,接着可以继续放入左括号,右括号放入的条件是左括号数目多于右括号,其余情况直接退出即可。
class Solution:
def generateParenthesis(self, n):
"""
:type n: int
:rtype: List[str]
"""
rr=[]
s=''
self.help(rr, 0, 0, s, n)
return rr
def help(self, rr, opennum, closenum, s, n):
if len(s) == 2*n:
rr.append(s)
return
if opennum < n:
self.help(rr, opennum+1, closenum, s+'(', n)
if closenum < opennum:
self.help(rr, opennum, closenum+1, s+')', n)
判断一个括号序列是否是匹配的:
def defer(s='(()()()'):
r=[]
for c in s:
if c=='(':
r.append(c)
else:
if len(r)==0 or (len(r)>1 and r[-1]!='('):
return False
r.pop(-1)
if len(r)==0:
return True
else:
return False