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('(',…

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…

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…

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…

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…

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: "((()))", "(()())", "(())()", "()(())", "()()()&…

[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…

# 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…