22 bracket generation

One 、 Preface

label ： Backtracking algorithm .

Source of problem LeetCode 22 difficulty ： secondary .

Question link ：https://leetcode-cn.com/problems/generate-parentheses/

 

Two 、 subject

Numbers  n  Represents the logarithm of the generated bracket , Please design a function , Used to be able to generate all possible and   Effective   Bracket combination .

Example ：

 Input ：n = 3
 Output ：[
       "((()))",
       "(()())",
       "(())()",
       "()(())",
       "()()()"
     ]

 

3、 ... and 、 Ideas

Backtracking solution , May refer to 《46 Total sort 》

 

Four 、 coded

//==========================================================================
/*
* @file    : 022_GenerateParenthesis.h
* @label   :  Backtracking algorithm 
* @blogs   : https://blog.csdn.net/nie2314550441/article/details/107573657
* @author  : niebingyu
* @date    : 2020/07/25
* @title   : 22. Bracket generation 
* @purpose :  Numbers  n  Represents the logarithm of the generated bracket , Please design a function , Used to be able to generate all possible and   Effective   Bracket combination .
*
*  Example ：
*  Input ：n = 3
*  Output ：[
*        "((()))",
*        "(()())",
*        "(())()",
*        "()(())",
*        "()()()"
*      ]
*
*
*  source ： Power button （LeetCode）
*  difficulty ： secondary 
*  link ：https://leetcode-cn.com/problems/generate-parentheses/
*/
//==========================================================================
#pragma once
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
#include <assert.h>
using namespace std;

#define NAMESPACE_GENERATEPARENTHESIS namespace NAME_GENERATEPARENTHESIS {
#define NAMESPACE_GENERATEPARENTHESISEND }
NAMESPACE_GENERATEPARENTHESIS

class Solution
{
public:
    vector<string> generateParenthesis(int n)
    {
        vector<string> result;
        string current;
        backtrack(result, current, 0, 0, n);
        return result;
    }

private:
    void backtrack(vector<string>& ans, string& cur, int open, int close, int n)
    {
        if (cur.size() == n * 2)
        {
            ans.push_back(cur);
            return;
        }

        if (open < n)
        {
            cur.push_back('(');
            backtrack(ans, cur, open + 1, close, n);
            cur.pop_back();
        }

        if (close < open)
        {
            cur.push_back(')');
            backtrack(ans, cur, open, close + 1, n);
            cur.pop_back();
        }
    }
};


 Here is the test code //
//  test   Use cases  START
void test(const char* testName, int n, vector<string> expect)
{
    Solution s;
    vector<string> result = s.generateParenthesis(n);

    if (result == expect)
        cout << testName << ", solution passed." << endl;
    else
        cout << testName << ", solution failed. " << endl;
}

//  The test case 
void Test1()
{
    int n = 3;
    vector<string> expect =
    {
        "((()))",
        "(()())",
        "(())()",
        "()(())",
        "()()()"
    };

    test("Test1()", n, expect);
}

NAMESPACE_GENERATEPARENTHESISEND
//  test   Use cases  END
//
void GenerateParenthesis_Test()
{
    cout << "------ start 22. Bracket generation  ------" << endl;
    NAME_GENERATEPARENTHESIS::Test1();
    cout << "------ end 22. Bracket generation  --------" << endl;
}

Execution results ：