LeetCode || Generate Parentheses
2014-04-10 22:26
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Generate Parentheses
Total Accepted: 9845 TotalSubmissions: 32228My Submissions
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
思路:典型的递归问题,递归终止的条件是1、左括号数量少于右括号;2、左括号数量多于n个;3、当左括号==右括号数量,且个数为n时,生成一个结果。
class Solution { public: vector<string> generateParenthesis(int n) { vector<string> res; func(res, n, 0, 0, string()); return res; } private: void func(vector<string> &res, int n, int left, int right, string temp){ if(left>n||left<right) return; if(left==right&&left==n){ res.push_back(temp); return; } func(res, n, left+1, right, temp+"("); func(res, n, left, right+1, temp+")"); } };
经验:递归问题要理解递归的过程,是一个不断压栈、弹栈的过程,比如上面最后两句递归调用的代码,虽然是先添加“(”,再添加“)”,表面上看,总是一直加入左括号,然后才加入右括号,但实际上不是这样,每次第一个递归返回时,紧接着进行第二次递归调用时,就是弹出“(”,先加入“)”的过程,而不是一直添加“(”直至达到数量n,才开始添加“)”。
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