标签:
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:递归:也就两种选择,要么左括号,要么右括号,分别记录个数;只要左括号的个数大于0的话,那么就可以填写左括号;右括号的话,那么要确保此时的序列,左括号的个数要大于右括号的个数。
class Solution { public: void print(int left, int right, string s, vector<string> &result) { if (left == 0 && right == 0) { result.push_back(s); } if (left > 0) { print(left-1, right, s+'(', result); } if (right > 0 && left < right) { print(left, right-1, s+')', result); } } vector<string> generateParenthesis(int n) { vector<string> result; string tmp = ""; print(n, n, tmp, result); return result; } };
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原文地址:http://blog.csdn.net/u011345136/article/details/43669955