[leetcode+DFS] DFS的常见写法 / 22. Generate Parentheses

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 + ')')