参考资料:https://www.youtube.com/watch?v=sz1qaKt0KGQ
DFS总结：
一般DFS： def fun() DFS() def DFS() if: DFS 带回溯的DFS： def fun() DFS() def DFS() if: ret.append() DFS onecase.pop()
#注意递归里要写成这样 if left > 0:self.DFS(ret, n, left -1, right, onecase + '(') if right > 0:self.DFS(ret, n, left, right -1, onecase + ')') #不是这样 这样会丢失中间子树的右节点。 if left > 0: left -= 1 onecase = onecase + '(' self.DFS(ret, n, left, right, onecase) while right > 0: #if will lose some right branch such as '(())' right -= 1 onecase = onecase + ')' self.DFS(ret, n, left, right, onecase)
回到题目上：
树生成策略：
）不能位于开头 [left == n right =n -1]
只有在前面有（ 没有被匹配的时候（left大于right）的时候 才能加上 ）[if left - right >0 append(')']
不能只剩下（ （left大于零 right 等于0）[left >0 right ==0]
当left==0而right>0时，把所有的右括号都加上然后递归
盯着题目发了半小时呆，决定去查U2B。真是会了5分钟不会半小时啊.
Runtime:
64 ms, faster than 10.05% of Python3 online submissions forGenerate Parentheses.
Memory Usage:
13 MB, less than 90.88% of Python3 online submissions forGenerate Parentheses.
Submission Detail
|
8 / 8 test cases passed.
|
Status: Accepted |
|
Runtime: 64 ms
Memory Usage: 13 MB
|
Submitted: 0 minutes ago
|
class Solution: def generateParenthesis(self, n: int) -> List[str]: ret = [] # 0 if n == 0: return ret # 1 if n == 1: ret.append('()') return ret # general # left:number of '(' right:number of ')' onecase = '' self.DFS(ret, n, n, n, onecase) return ret def DFS(self, ret, n, left, right, onecase): # #cut tree if (left ==n and right ==n-1) or (left >0 and right ==0) or (left > right): return #ret ahead if left ==0 and right >0: onecase = onecase + ')'*right right = 0 if ret.count(onecase) == 0: ret.append(onecase) #ret if left ==0 and right ==0: if ret.count(onecase) ==0: ret.append(onecase) # if left > 0:self.DFS(ret, n, left -1, right, onecase + '(') if right > 0:self.DFS(ret, n, left, right -1, onecase + ')')
ver2:
Runtime:
36 ms, faster than 98.00% of Python3 online submissions forGenerate Parentheses.
Memory Usage:
13.3 MB, less than 71.53% of Python3 online submissions for Generate Parentheses.
Submission Detail
|
8 / 8 test cases passed.
|
Status: Accepted |
|
Runtime: 36 ms
Memory Usage: 13.3 MB
|
Submitted: 5 minutes ago
|
class Solution:
def generateParenthesis(self, n: int) -> List[str]:
#0
if n ==0:
return []
#1
if n==1:
return["()"]
#normal
ret = []
onecase = ''
self.DFS(ret,n,n,onecase)
return ret
def DFS(self,ret,left,right,onecase):
if left > right:
return
if left ==0 and right ==0:
ret.append(onecase)
if left >0:self.DFS(ret,left -1,right,onecase + '(')
if right >0:self.DFS(ret,left,right -1,onecase + ')')