一 顺时针打印矩阵

class Solution {public:    vector
     
       printMatrix(vector
      
       
         > matrix) {        vector
        
          result;        int m = matrix.size();        if (m==0) return result;        int n = matrix[0].size();        if (n==0) return result;        int start=0;        while(start < m/2.0 && start < n/2.0) {           printCircle(matrix, start, m, n, result);           start++;        }        return result;    }// 先判断问题，拆成print Circle, // 确定起点和终点的变化（看坐标，起点和终点的坐标相加和相同）// 四个顺序，除了第一个，剩下的都是有条件的，并且要执行后一个必须要满足前面所有的条件。private:    void printCircle(vector
         
          
           > matrix, int start, int m, int n, vector
           
             & result){ int end_x = n - 1 - start; int end_y = m - 1 - start; // 从左到右 for (int i = start; i <= end_x; i++) result.push_back(matrix[start][i]); // 从上到下 if (end_y > start) { for (int i = start+1; i <= end_y; i++) result.push_back(matrix[i][end_x]); } // 从右到左 if (end_x - 1 >= start && end_y > start) { for (int i= end_x-1; i>=start; i--) result.push_back(matrix[end_y][i]); } // 从下到上 if (end_y - 1 > start && end_x -1 >= start) { for (int i= end_y - 1; i>= start+1; i--) result.push_back(matrix[i][start]); } }};