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))
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持【听图阁-专注于Python设计】。
