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：
有关Parentheses的题目也是比较常见的，除了Generate Parentheses还有Valid Parentheses，Longest Valid Parentheses
回溯用递归写是性价比最高的，基本思想就是从前往后填字符保证'('的个数不能比')'少，且'('的个数不能比n多
代码：
void generateAtDepth(vector<string> &Parentheses, string &p, int lefts, int rights, int depth, int n){
if(depth == *n-){
p[depth] = ')';
Parentheses.push_back(p);
return;//返回值为void的函数不能忘了return啊！
}
if(lefts < n){//注意不能出现((()
p[depth] = '(';
generateAtDepth(Parentheses, p, lefts+, rights, depth+, n);
}
if(lefts > rights){
p[depth] = ')';
generateAtDepth(Parentheses, p, lefts, rights+, depth+, n);
}
}
vector<string> generateParenthesis(int n) {
vector<string> Parentheses;
if(n < ) return Parentheses;
string p(*n, '*');//一定要指定初始化字符
generateAtDepth(Parentheses, p, , , , n);
return Parentheses;
}
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