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：
DFS.
void addParenthesis(vector<string>& res, string str, int open, int close,int n) { if (str.length() == n*2) { res.push_back(str); return; } if (open<n)addParenthesis(res, str + "(", open + 1, close, n); if (close < open)addParenthesis(res, str + ")", open, close + 1, n); } vector<string> generateParenthesis(int n) { vector<string>res; addParenthesis(res, "", 0, 0, n); return res; }
参考：
https://discuss.leetcode.com/topic/4485/concise-recursive-c-solution
[leetcode-22-Generate Parentheses]
原文：http://www.cnblogs.com/hellowooorld/p/6909946.html