Question :
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
Anwser 1 :
class Solution { public: vector<string> generateParenthesis(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<string> vec; string str; addParen(vec,str, n, n); return vec; } void addParen(vector<string> &vec, string str, int left, int right) { if (left == 0 && right == 0) { vec.push_back(str); return; } if (left > 0) { addParen(vec, str + '(', left - 1, right); } if (right > left) { addParen(vec, str + ')', left, right-1); } } };
Anwser 2 :
class Solution { public: void printPar(int l, int r, vector<string>& result, char* str, int idx) { if (l < 0 || r < l) return; if (l == 0 && r == 0) { str[idx] = '\0'; result.push_back(string(str)); } else { if (l > 0) { str[idx] = '('; printPar(l-1, r, result, str, idx+1); } if (r > l) { str[idx] = ')'; printPar(l, r-1, result, str, idx+1); } } } vector<string> generateParenthesis(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<string> result; char* str = new char[2*n+1]; printPar(n, n, result, str, 0); delete []str; return result; } };
Anwser 3 :
class Solution { public: vector<string> generateParenthesis(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<string> ans; if (n > 0) { generator(ans, "", 0, 0, n); } return ans; } // r/l: appearance of ) ( void generator(vector<string> & ans, string s, int l, int r, int n) { if (l == n) { ans.push_back(s.append(n-r, ')')); return; } generator(ans, s + '(', l+1, r, n); if (l > r) { generator(ans, s + ")", l, r+1, n); } } };
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