Givennpairs of parentheses, write a function togenerate all combinations of well-formed parentheses.
For example, givenn= 3, a solution set is:
"((()))","(()())", "(())()", "()(())", "()()()"
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#pragma once#include<iostream>#include<sstream>#include<string>#include<vector>using namespace std;//将数组temp中字符合成字符串push进resultvoid pushto(char* temp, int n, vector<string> &result){string s;stringstream ss;for (int i = 0; i < n; i++)ss << temp[i];result.push_back(ss.str());}//now:temp当前要赋值的位置void generateandpush(char *temp, vector<string> &result,int now,int n,int count){if (count < 0)//此路不通return;if (now == n){if (count==0)//最后，，净左括号数应为0pushto(temp, n, result);return;}temp[now] = '(';count++;generateandpush(temp, result, now + 1, n,count);count–;//恢复temp[now] = ')';count–;generateandpush(temp, result, now + 1, n,count);return;}vector<string> generateParenthesis(int n){char *temp = new char[2*n];//n对vector<string> result;int count = 0;//记录 左括号-右括号generateandpush(temp, result,0,2*n,count);return result;}void main(){vector<string> result;result = generateParenthesis(3);for (int i = 0; i < result.size(); i++)//printf("%s", result[0].c_str());//注意printf打印string时要用：s.c_str()cout << result[i] << ' ';cout << endl;system("pause");}
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