LeetCode 21. Generate Parentheses
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参考了remlostime的题解
将返回值ret的引用作为传出参数,递归产生括号,并加上剪枝:left_tot*2 > max_dep时,即当前压入的左括号的数量(left_tot*2 > 2*n = max_dep=> left_tot > n)超过总数n时剪枝即可。
代码:
class Solution {public: vector<string> generateParenthesis(int n) { vector<string> ret; gao(0, 2*n, 0, 0, string(), ret); return ret; }private:void gao(int dep, int max_dep, int left_now, int left_tot, string s, vector<string>& ret){if (left_tot*2 > max_dep){return ;} else if (dep == max_dep){ret.push_back(s);return ;}gao(dep+1, max_dep, left_now+1, left_tot+1, s+'(', ret);if (left_now > 0){gao(dep+1, max_dep, left_now-1, left_tot, s+')', ret);}}};
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