输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字 C++实现

题目描述：

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字，例如，如果输入如下矩阵： 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.

int main()

{

         Solution a;

         vector<vector<int>>matrix = { {1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15} };//矩阵初始化

         vector<int> m=a.printMatrix(matrix);

         for(autoi : m)//依次打印返回矩阵的值

                   cout<<i;

         return0;

}

方法一：

//每次循环找出右上和左下的点作为参考

class Solution{

public:

         vector<int>printMatrix(vector<vector<int>>  matrix) {

                   vector<int>vec;

                   int row = matrix.size();

                   int column = matrix[0].size();

                   int count = 0;

                   if(row == 1)//区分行向量和列向量

                            vec=matrix[0];

                   elseif(column == 1)

                            {

                                     for(int i = 0; i < row; i++)

                                     vec.push_back(matrix[i][0]);

                            }

                   else{

                   for(int m = 0;(m < row - count)&&(m<column-count); m++)

                   {

                            count++;

                            if(column- 1 - m>=m){

                            int right_up = matrix[m][column- 1 - m];

                            for(int i = m; i < column - 1 - m; i++) vec.push_back(matrix[m][i]);

                            vec.push_back(matrix[m][column- 1-m]);//右上元素

                            for(int j = m+1; j < row - m-1; j++) vec.push_back(matrix[j][column-1- m]);}

                            if(row - 1 - m >m)

                            {

                            for(int i = column - 1 - m; (i > m); i--) vec.push_back(matrix[row- 1 - m][i]);

                            vec.push_back(matrix[row- 1 - m][m]);//左下元素

                            for(int i = row - 1 - m-1; i > m; i--) vec.push_back( matrix[i][m]);

                            }

                   }

                   }

                   return vec;

         }

};

方法二：

//直接定义一个矩形，在矩形的四条边取值，程序大大简化

class Solution{

public:

         vector<int>printMatrix(vector<vector<int>>  matrix) {

                   vector<int>vec;

                   int row = matrix.size();

                   int column = matrix[0].size();

                   int left = 0, right = column, up = 0, bottom = row;

                   while((up < bottom) && (left < right))

                   {

                            for(int i = left; i < right; i++) vec.push_back(matrix[up][i]);

                            for(int i = up+1; i < bottom; i++) vec.push_back(matrix[i][right-1]);

                            for(int i = right-1-1; ((bottom-1)!=up)&&(i >=left); i--) vec.push_back(matrix[bottom-1][i]);

                            for(int i = bottom-1-1; ((right-1)!=left)&&(i >left); i--) vec.push_back(matrix[i][left]);

                            up++;left++; right--; bottom--;

                   }

                   return vec;

         }

};