22. Generate Parentheses(ML)
22. Generate Parentheses
. Generate Parentheses Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = , a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]
class Solution(object):
def generateParenthesis(self, n):
"""
:type n: int
:rtype: List[str]
"""
if n == 0: return ['']
ans = []
for c in xrange(n):
for left in self.generateParenthesis(c):
for right in self.generateParenthesis(n-1-c):
ans.append('({}){}'.format(left, right))
return ans
class Solution(object):
# Brute Force
def generateParenthesis(self, n):
def generate(A=[]):
if len(A) == 2 * n:
if valid(A):
ans.append("".join(A))
else:
A.append('(')
generate(A)
A.pop()
A.append(')')
generate(A)
A.pop()
def valid(A):
bal = 0
for c in A:
if c == '(':
bal += 1
else:
bal -= 1
if bal < 0: return False
return bal == 0
ans = []
generate()
return ans
# Backtracking
def generateParenthesis2(self, N):
ans = []
def backtrack(S='', left=0, right=0):
if len(S) == 2 * N:
ans.append(S)
return
if left < N:
backtrack(S + '(', left + 1, right)
if right < left:
backtrack(S + ')', left, right + 1)
backtrack()
return ans
# Closure Number
def generateParenthesis3(self, N):
if N == 0: return ['']
ans = []
for c in range(N):
# for c in xrange(N):
for left in self.generateParenthesis(c):
for right in self.generateParenthesis(N - 1 - c):
ans.append('({}){}'.format(left, right))
return ans
sn = Solution()
print(sn.generateParenthesis(3))
print(sn.generateParenthesis2(3))
print(sn.generateParenthesis3(3))




22. Generate Parentheses(ML)的更多相关文章
- [Leetcode][Python]22: Generate Parentheses
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 22: Generate Parentheseshttps://oj.leet ...
- 刷题22. Generate Parentheses
一.题目说明 这个题目是22. Generate Parentheses,简单来说,输入一个数字n,输出n对匹配的小括号. 简单考虑了一下,n=0,输出"";n=1,输出" ...
- 【LeetCode】22. Generate Parentheses (2 solutions)
Generate Parentheses Given n pairs of parentheses, write a function to generate all combinations of ...
- 22. Generate Parentheses (recursion algorithm)
Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ...
- [LeetCode] 22. Generate Parentheses 生成括号
Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ...
- LeetCode 22. Generate Parentheses
Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ...
- 22. Generate Parentheses——本质:树,DFS求解可能的path
Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ...
- Java [leetcode 22]Generate Parentheses
题目描述: Given n pairs of parentheses, write a function to generate all combinations of well-formed par ...
- leetcode@ [22]Generate Parentheses (递归 + 卡特兰数)
https://leetcode.com/problems/generate-parentheses/ Given n pairs of parentheses, write a function t ...
随机推荐
- 一、kubeadm安装
一.官网 https://kubernetes.io/zh/docs/setup/independent/install-kubeadm/ 阿里云 kubernetes yum 仓库镜像 安装kube ...
- 二、.Net 连接mycat
一.mycat 单体的mysql已经过去 二.引用Mycat包 三.代码 using Pomelo.Data.MyCat; using System; using System.Collections ...
- codeforces749B
Parallelogram is Back CodeForces - 749B 已知平行四边形的三个顶点,求第四个顶点可能的位置.Input输入有三行,每行包括两个整数x和y ( - 1000 ≤ x ...
- linux-shell系列3-wafAPI
#!/bin/bash datestr=`env LANG=en_US.UTF-8 date -u "+%a, %d %b %Y %H:%M:%S GMT"`pwdstr=`ech ...
- 浅析Android设备中grep命令处理流程
2017-04-18 概述 在TV开发板中,可以在串口中直接使用grep命令.这是因为在/system/bin/下有一个'grep'链接.这个链接指向'/system/bin/toolbo ...
- IDEA Failed to prepare an update: Temp directory inside installation
具体错误: Connection Error Failed to prepare an update: Temp directory inside installation: F:\IDEA_Tool ...
- 百度分享不支持https的解决方案(单独部署静态文件)
首先是参考了博客,下载百度分享的静态代码 static 链接为:https://www.cnblogs.com/mmzuo-798/p/6434576.html 后来在nginx的 nginx.con ...
- 【POJ1037】A decorative fence(DP)
BUPT2017 wintertraining(15) #6C 题意 给长度n的数列,1,2,..,n,按依次递增递减排序,求字典序第k小的排列. 题解 dp. up[i][j]表示长度为j,以第i小 ...
- poj1845 sumdiv (因数的和)
首先分解质因数,$A^B=p_1^{m_1B}p_2^{m_2B}...p_n^{m_nB}$ 然后的话,它的所有因数的和就是$\prod{(1+p_i^1+p_i^2+...+p_i^n)}$ 用一 ...
- [WC2008]游览计划(状压dp)
题面太鬼畜不粘了. 题意就是给一张n*m的网格图,每个点有点权,有k个关键点,让你把这k个关键点连成一个联通快的最小代价. 题解 这题nmk都非常小,解法肯定是状压,比较一般的解法插头dp,但不太好写 ...