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:
"((()))", "(()())", "(())()", "()(())", "()()()"
#include <vector>
#include <string>
#include <iostream>
using namespace std;
void generateParenthesis(int left, int right, vector<string>& res, string path) {
if(left == 0 && right == 0) {
res.push_back(path);
return;
}
if(left > 0)
generateParenthesis(left-1, right, res, path + '(');
if(right > left)
generateParenthesis(left, right-1, res, path + ')');
}
vector<string> generateParenthesis(int n) {
vector<string> res;
string path = "";
generateParenthesis(n, n, res, path);
return res;
}
}
int main(void) {
vector<string> res = generateParenthesis(3);
for(int i = 0; i < res.size(); ++i) {
cout << res[i] << endl;
}
}
本文介绍了一个递归算法,用于生成所有合法的n对括号组合。通过递归地添加左括号和右括号,并确保在任何时候都不出现非法的括号序列,该算法能有效地生成所有可能的组合。
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