题意: 产生n对合法括号的所有组合,用vector<string>返回. 思路: 递归和迭代都可以产生.复杂度都可以为O(2n*合法的括号组合数),即每次产生出的括号序列都保证是合法的. 方法都是差不多的,就是记录当前产生的串中含有左括号的个数cnt,如果出现右括号,就将cnt--.当长度为2*n的串的cnt为0时,就是答案了,如果当前cnt比剩下未填的位数要小,则可以继续装“(”,否则不能再装.如果当前cnt>0,那么就能继续装“)”与其前面的左括号匹配(无需要管匹配到谁,总之能匹配)…

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: "((()))", "(()())", "(())()", "()(())", "()()()" 在LeetCo…

9.6 Implement an algorithm to print all valid (e.g., properly opened and closed) combinations of n-pairs of parentheses.EXAMPLEInput: 3Output: ((())), (()()), (())(), ()(()), ()()() LeetCode上的原题,请参见我之前的博客Generate Parentheses 生成括号. 解法一: class Solution…

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]"…

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. Have you met this question in a real interview?     Example Given n = 3, a solution set is: "((()))", "(()())", "(())()",…

回溯法 百度百科:回溯法(探索与回溯法)是一种选优搜索法,按选优条件向前搜索,以达到目标.但当探索到某一步时,发现原先选择并不优或达不到目标,就退回一步又一次选择,这样的走不通就退回再走的技术为回溯法,而满足回溯条件的某个状态的点称为"回溯点". 在包括问题的全部解的解空间树中,依照深度优先搜索的策略,从根结点出发深度探索解空间树.当探索到某一结点时,要先推断该结点是否包括问题的解,假设包括,就从该结点出发继续探索下去,假设该结点不包括问题的解,则逐层向其祖先结点回溯.(事实上回溯法就…

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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] 给定一…

Generate ParenthesesGiven n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "…

Generate Parentheses: 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: [ "((()))", "(()())", "(())()", "()(())", &q…

Given a string containing just the characters'(',')','{','}','['and']', determine if the input string is valid. The brackets must close in the correct order,"()"and"()[]{}"are all valid but"(]"and"([)]"are not. 题意:给…

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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]   …

题意 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: [ "((()))", "(()())", "(())()", "()(())", "()()()"] 输…

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: "((()))", "(()())", "(())()", "()(())", "()()()" 求出所有可能的…

描述 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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]…

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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] 题意:…

问题解法参考 它给出了这个问题的探讨. 超时的代码: 这个当n等于7时,已经要很长时间出结果了.这个算法的复杂度是O(n^2). #include<iostream> #include<vector> #include<stack> #include<map> using namespace std; bool isValid(string s) { map<char, char> smap; smap.insert(make_pair('(',…

[称号] 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: "((()))", "(()())", "(())()", "()(())", "()()()" [题…

给 n 对括号,写一个函数生成所有合适的括号组合.比如,给定 n = 3,一个结果为:[  "((()))",  "(()())",  "(())()",  "()(())",  "()()()"]详见:https://leetcode.com/problems/generate-parentheses/description/ 由于字符串只有左括号和右括号两种字符,而且最终结果必定是左括号3个,右括号3个…

题目 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:  [   "( ( ( ) ) )",   "( ( ) ( ) )",   "( ( ) ) ( )",   "( ) ( ( ) )&q…

# 解题思路:列举出所有合法的括号匹配,使用dfs.如果左括号的数量大于右括号的数量的话,就不能产生合法的括号匹配class Solution(object): def generateParenthesis(self, n): """ :type n: int :rtype: List[str] """ if n == 0: return [] res = [] self.recursion(n,n,'',res) return res def…

Description: 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 { public: void push(char c){ //插入结点 struct node *n=new struct node; n->nex=; n->ch=c; n->pre=last; last->nex=n; last=last->nex; } bool jud(char c){ //判断 struct node *m; if(c==']') c='['; else if(c=='}') c='{'; else if(c==')')…

2道题目都差不多,就是问和相邻所有点都有相同数据相连的作为一个联通快,问有多少个连通块 因为最近对搜索题目很是畏惧,总是需要看别人代码才能上手,就先拿这两道简单的dfs题目来练练手,顺便理一理dfs的思路,分析清楚dfs的退出递归的条件和什么时候进行递归调用是至关重要的,这两道题目不涉及回溯,对于需要回溯的题目也要清楚分析,找到回溯条件,在对一个新的状态dfs时,后面加上回溯的语句 HDU1241代码: #include <cstdio> #include <cstring> #i…

Test1.A Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 sdut 2274:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2274 给定一个2行n列的棋盘,从上到下行号为1..2,从左到右列号为1..n.有些格子可走,有些格子不可走.可走的格子用'.'表示,不可走的格子用'W'表示.现在路人甲想从格子(1,1)走到格子(1,…

题目链接: https://vjudge.net/problem/POJ-3009 题目描述: On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked…

Leetcode之回溯法专题-22. 括号生成(Generate Parentheses) 给出 n 代表生成括号的对数,请你写出一个函数,使其能够生成所有可能的并且有效的括号组合. 例如,给出 n = 3,生成结果为: [ "((()))", "(()())", "(())()", "()(())", "()()()"] 分析:给定一个n,生成所有可能的括号组合. 举个例子,n=3,需要生成三个括号,那最…

22. 括号生成 22. Generate Parentheses 题目描述 给出 n 代表生成括号的对数,请你写出一个函数,使其能够生成所有可能的并且有效的括号组合. 例如,给出 n = 3,生成结果为: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] LeetCode22. Generate Parentheses中等回溯算法 Java 实现 i…

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, -and *. Example 1 Input: "2-1-1". ((2-1)-1) = 0 (2-(1-1)) = 2 Output: …

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *. Example 1: Input: "2-1-1" Output: [0, 2] Explanation: ((2-1)-…

目录 题目链接 注意点 解法 小结 题目链接 Generate Parentheses - LeetCode 注意点 解法 解法一:递归.当left>right的时候返回(为了防止出现 )( ) class Solution { public: void recursion(int left,int right,string str,vector<string> &ret) { if(left > right) return; else if(left== 0&&a…