白菜刷LeetCode记-22. Generate Parentheses 今天的题目如下: 这道题目是给出n对圆括号,写一个函数去生成由所有合法匹配的组合生成的数组。 首先想到的是将所有的组合罗列出来,并分别校验组合是否合法,但是这种方法可能就会导致比较不好的时间复杂度了。那么就要想想什么情况会是合法的确情况了。 假设左括号数为 lnum,
leetcode-mid-backtracking -22. Generate Parentheses-NO mycode 没有思路,大早上就有些萎靡,其实和上一个电话号码的题如出一辙啦 参考: classSolution(object):defgenerateParenthesis(self, n):""":type n: int :rtype: List[str]"""defdfs(temp,l,r):ifl == 0andr ==0: res.append(temp) #retur...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
[Leetcode] Generate Parentheses Givennpairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, givenn= 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()" Solution: 1publicclassSolution {2publicList<S...
[leetcode] generate parentheses Generate Parentheses Givennpairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, givenn= 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()"...
把左右括号剩余的次数记录下来,传入回溯函数。 判断是否得到结果的条件就是剩余括号数是否都为零。 注意判断左括号是否剩余时,加上left>0的判断条件!否则会memory limited error! 判断右括号时要加上i==1的条件,否则会出现重复的答案。 同样要注意在回溯回来后ans.pop_back() ...
[LeetCode 题解]: Generate Parentheses Givennpairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, givenn= 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()"...
注意到卡塔兰数的递推性质: an= an-1a0+ an-2a1+ ... + an-2a1+ an-1a0。 1classSolution2{3public:4vector<string> generateParenthesis(intn)5{6vector<vector<string>> table(n +1);7table[0] = vector<string>({""});8inti, j, k, lp;9for(i =1; i <= n; i++)//write all...
LeetCode 22 Generate Parentheses Givennpairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, givenn= 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()"...
题目链接:https://leetcode.com/problems/generate-parentheses/description/ 题目大意:给出一个数字n,找出所有包含n对()括号对的组合,例子如下: 法一:直接用深搜,将(和)交换着加入string中,然后每得到一个string用stack判断是否是匹配的。代码如下(耗时57ms): ...