Improve spiral order

[Jesús Jiménez Sánchez]

The problem added in #17 contains the solution I presented in the interview. It can be improved because this solution would only work with a matrix of the same dimensions as the given ones. It should work with every type of matrix.

This is the implementation at the moment:

Interview-problems/spiral_order/spiral_order.py

Lines 19 to 87 in 84b1292

 three_by_three = [ 
    [1, 2, 3], 
    [4, 5, 6], 
    [7, 8, 9], 
 ] 
  
 five_by_two = [ 
     [1, 2, 3, 4, 5], 
     [6, 7, 8, 9, 10] 
 ] 
  
 one_by_five = [ 
     [1], 
     [2], 
     [3], 
     [4], 
     [5], 
 ] 
  
  
 def spiral_order(matrix): 
     """ 
         Transform matrix into liist inn spiral order. 
  
         Parameters 
         ---------- 
         matrix : List(List()) 
             Matrix to be transformed to spiral form. 
     """ 
     result = [] 
     reversed_last_row = matrix[-1] 
     reversed_matrix = list(matrix) 
     saved_i = 0 
  
     reversed_last_row.reverse() 
     reversed_matrix.reverse() 
  
     for elem in matrix[0]: 
         result.append(elem) 
  
     result.pop() 
  
     for elem in matrix: 
         result.append(elem[-1]) 
  
     if len(matrix[-1]) > 1: 
         result.pop() 
  
         for elem in reversed_last_row: 
             result.append(elem) 
  
         if len(matrix) > 2: 
             for i in range(1, len(matrix) - 1): 
                 result.append(reversed_matrix[i][0]) 
                 saved_i = i 
  
             result.append(matrix[saved_i][1]) 
  
     return result 
  
  
 print(spiral_order(three_by_three)) 
 # [1, 2, 3, 6, 9, 8, 7, 4, 5] 
  
 print(spiral_order(five_by_two)) 
 # [1, 2, 3, 4, 5, 10, 9, 8, 7, 6] 
  
 print(spiral_order(one_by_five)) 
 # [1, 2, 3, 4, 5]