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：
不知道DFS应该写什么。其实就是把dfs代入不同的数值，再写一遍就行了。不同条件的可以分开写
dfs字符串的题目，一般有个String currentString。这里要判断左右开口数，还要加open close变量
class Solution { public List<String> generateParenthesis(int n) { List<String> results = new ArrayList<>(); //cc if (n < 0) return results; //dfs dfs(n, 0, 0, "", results); //return return results; } public void dfs(int n, int open, int close, String cur, List<String> results) { //exit if (cur.length() >= 2 * n) { results.add(cur); return ; } if (open < n) dfs(n, open + 1, close, cur + ‘(‘, results); if (close < open) dfs(n, open, close + 1, cur + ‘)‘, results); } }
22. Generate Parentheses产生所有匹配括号的方案
原文：https://www.cnblogs.com/immiao0319/p/13388355.html