python 实现矩阵按对角线打印

Description：
将一个矩阵（二维数组）按对角线向右进行打印。（搜了一下发现好像是美团某次面试要求半小时手撕的题）
Example:
Input:
[
[1,2,3,4],
[5,1,2,3],
[9,5,1,2]
]
Output:
[[4],
[3, 3],
[2, 2, 2],
[1, 1, 1],
[5, 5],
[9]]

(0,3)
(0,2)
(0,1)
(0,0)
(1,0)
(2,0)

[(0,3)]
[(0,2),(1,3)]
[(0,1),(1,2),(2,3)]
[(0,0),(1,1),(2,2)]
[(1,0),(2,1)]
[(2,0)]

def diagonal_right(matrix):
 if not matrix:
  return []
 row = len(matrix)
 col = len(matrix[0])
 col2 = col
 result = []
 for i in range(row):
  for j in range(col2 - 1, -1, -1): #j倒序遍历
   lst = []
   i1,j1 = i,j #i1,j1用于方便同一对角线元素的添加，否则改变i,j影响开头元素的选择
   while i1 <= row - 1 and j1 <= col - 1:
    lst.append(matrix[i1][j1])
    j1 += 1
    i1 += 1
   result.append(lst)
   if i == 0 and j == 0:#当遍历完(0,0)开头的一条对角线后，让j固定为0
    col2 = 1
 return('the result is: %s'%result,'end')
 
#test
print(diagonal_right([[1],[2]]))
print(diagonal_right([[1]]))
print(diagonal_right([
 [1,2,3,4],
 [5,1,2,3],
 [9,5,1,2]
]))

(0,0)
(0,1)
(0,2)
(0,3)
(1,3)
(2,3)

[(0,0)]
[(0,1),(1,0)]
[(0,2),(1,1),(2,0)]
[(0,3),(1,2),(2,1)]
[(1,3),(2,2)]
[(2,3)]

def diagonal_left(matrix):
 if not matrix:
  return []
 row = len(matrix)
 col = len(matrix[0])
 k=0
 result = []
 for i in range(row):
  for j in range(k,col): # j顺序遍历
   lst = []
   i1, j1 = i, j # i1,j1用于方便同一对角线元素的添加，否则改变i,j影响开头元素的选择
   while i1 <= row - 1 and j1 >=0:
    lst.append(matrix[i1][j1])
    j1 -= 1
    i1 += 1
   if i==0 and j==col-1:
    k=col-1
   result.append(lst)
 return ('the result is: %s' % result, 'end')

# test
print(diagonal_left([[1],[2]]))
print(diagonal_left([[1]]))
print(diagonal_left([
 [1,2,3,4],
 [5,1,2,3],
 [9,5,1,2]
]))