python实现转圈打印矩阵

Python  /  管理员 发布于 7年前   187

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 res1def 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 matrixif __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))

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