python实现转圈打印矩阵

本文实例为大家分享了python实现转圈打印矩阵的具体代码，供大家参考，具体内容如下

#! conding:utf-8
__author__ = "hotpot"
__date__ = "2017/10/28 9:40"

def return_edge(matrix, start_col, end_col, start_row, end_row):
 if start_row == end_row:
   return matrix[start_row][start_col:end_col+1]
 elif end_col ==start_col:
   res = []

for i in range(start_row,end_row+1):
     res.append(matrix[i][end_col])
   return res
 else:
   res2 =[]
   res3 =[]
   res4=[]
   res1 = matrix[start_row][start_col:end_col+1]
   for i in range(start_row+1,end_row+1):
     res2.append(matrix[i][end_col])
   for i in range(end_col-1,start_col-1,-1):
     res3.append(matrix[end_row][i])
   for i in range(end_row-1,start_row,-1):
     res4.append(matrix[i][start_row])
   res1.extend(res2)
   res1.extend(res3)
   res1.extend(res4)
   return res1
def spiralOrder( matrix):
 if matrix:
   row = len(matrix)-1
   col = len(matrix[0])-1
   start_row = 0
   start_col = 0
   end_row = row
   end_col = col
   res =[]
   while start_col<=end_col and start_row <= end_row:
     res.extend(return_edge(matrix,start_col,end_col , start_row ,end_row))
     start_col+=1
     end_col-=1
     start_row+=1
     end_row-=1
   return res
 else:
   return matrix
if __name__ == '__main__':
 matrix = [[0 for i in range(3) ]for j in range(3)]
 num=1
 for m in range(len(matrix)):
   for n in range(len(matrix[0])):
     matrix[m][n]=num
     num+=1

print(spiralOrder( matrix))

来源：https://blog.csdn.net/hotpotbo/article/details/78374025