[ 프로그래머스 ] 행렬 테두리 회전하기

문제

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https://school.programmers.co.kr/learn/courses/30/lessons/77485

풀이

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주어진 2차원 배열을 그리드 형태로 구현하고, 주어진 값 주변 원소를 이동하는 문제으로, 단순하게 주어진 방식을 구현하여 정답을 구하면 된다.

#include <string>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> solution(int rows, int columns, vector<vector<int>> queries) 
{
    vector<int> answer;

    int element = 1;

    vector<vector<int>> matrix(rows);

    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < columns; j++)
        {
            matrix[i].push_back(element);
            element++;
        }
    }

    
    for (int i = 0; i < queries.size(); i++)
    {
        int temp, temp1;

        //2,2,5,4 = 2,2 -> 2,3 -> 2,4 -> 3,4 ->4,4 -> 5,4 -> 5,3 -> 5,2 -> 4,2 -> 3,2 -> 2,2 

        int m_row = queries[i][0] - 1;
        int m_colum = queries[i][1] - 1;
        int row_cnt = 0;
        int colum_cnt = 0;

        temp1 = matrix[m_row + row_cnt][m_colum + colum_cnt];

        int cnt = 0;

        vector<int> array;

        for (int j = 0; j < queries[i][3] - queries[i][1]; j++)
        {
            if (cnt % 2 == 0)
            {
                colum_cnt++;
                temp = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp1;

                array.push_back(temp1);
                cnt++;
            }

            else
            {
                colum_cnt++;
                temp1 = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp;

                array.push_back(temp);
                cnt++;
            }
        }

        for (int j = 0; j < queries[i][2] - queries[i][0]; j++)
        {
            if (cnt % 2 == 0)
            {
                row_cnt++;
                temp = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp1;

                array.push_back(temp1);
                cnt++;
            }

            else
            {
                row_cnt++;
                temp1 = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp;

                array.push_back(temp);
                cnt++;
            }
        }

        for (int j = 0; j < queries[i][3] - queries[i][1]; j++)
        {
            if (cnt % 2 == 0)
            {
                colum_cnt--;
                temp = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp1;

                array.push_back(temp1);
                cnt++;
            }

            else
            {
                colum_cnt--;
                temp1 = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp;

                array.push_back(temp);
                cnt++;
            }
        }

        for (int j = 0; j < queries[i][2] - queries[i][0]; j++)
        {
            if (cnt % 2 == 0)
            {
                row_cnt--;
                temp = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp1;

                array.push_back(temp1);
                cnt++;
            }

            else
            {
                row_cnt--;
                temp1 = matrix[m_row + row_cnt][m_colum + colum_cnt];
                matrix[m_row + row_cnt][m_colum + colum_cnt] = temp;

                array.push_back(temp);
                cnt++;
            }
        }

        int min = *min_element(array.begin(), array.end());

        answer.push_back(min);

    }

    return answer;
}