22. Generate Parentheses——本质:树,DFS求解可能的path
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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:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
class Solution(object): def generateParenthesis(self, n): """ :type n: int :rtype: List[str] [ "((()))", "(()())", "(())()", "()(())", "()()()"] () / () () / \ / () ()()() """ ans = [] path = [] self.gen_par_helper(n*2, path, ans) return ans def is_valid_par(self, par): stack = [] for c in par: if c == "(": stack.append("(") else: if stack: stack.pop() else: return False return len(stack) == 0 def gen_par_helper(self, n, path, ans): if n == 0: if self.is_valid_par(path): ans.append("".join(path)) return for c in "()": path.append(c) self.gen_par_helper(n-1, path, ans) path.pop()
class Solution(object): def generateParenthesis(self, n): """ :type n: int :rtype: List[str] [ "((()))", "(()())", "(())()", "()(())", "()()()"] () / () () / \ / () ()()() """ ans = [] self.gen_par_helper(n, n, "", ans) return ans def gen_par_helper(self, left, right, path, ans): if left == 0 and right == 0: ans.append(path) return if left > 0: self.gen_par_helper(left-1, right, path+"(", ans) if right > 0 and right > left: self.gen_par_helper(left, right-1, path+")", ans)
第二种更快,只是不那么容易想到!
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