21. Generate Parentheses Leetcode Python

leetcode全套解答python版本 leetcode全套解答python版本。包括更新到10月份的的leetcode 立即下载
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:


"((()))", "(()())", "(())()", "()(())", "()()()"

看到这道题目首先想到的是catlan Number  根据catlan number我们可以知道最后的解的数量是catlan number


这里要求我们求解我们可以用bfs枚举来解决。

1.当左括号数量小于n 时候添加左括号

2.当有括号数量小于左括号时候添加右括号

3.当总括号长度等于2n时候将解加入解的集合。


this is a catlan number problem. we can know the final solution length is the corresponding catlan number. 

to give specific solutions 

1. when # of left parentheses is no greater than n we append left parentheses

2. when # of right parentheses is not greater than left parentheses we append right parentheses

3. when the total length is 2n we append it to solution


code is as follow:

class Solution:
    # @param an integer
    # @return a list of string
    def findp(self,valuelist,solution,n,left,right):
        if len(valuelist)==2*n:
            solution.append(valuelist)
        if left<n:
            self.findp(valuelist+'(',solution,n,left+1,right)
        if right<left:
            self.findp(valuelist+')',solution,n,left,right+1)
    def generateParenthesis(self, n):
        solution=[]
        self.findp('',solution,n,0,0)
        return solution
        



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【LeetCode】22. Generate Parentheses 解题报告(Python)

题目分析: 这一题是让输入正整数n表示n对括号,然后输出括号所有匹配数,与【LeetCode】17. Letter Combinations of a Phone Number是有些相似的,因为可能的情况很多用循环没法解决,这时候考虑递归或动态规划,递归的代码比较简单,用递归便可解决。 代码说明: 1、def dfs(res, string, n, l, r):,递归函数的参数意义分别为,返回用的...

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