# 輸出遍歷的索引與其對應(yīng)的值
A = [[1,2,3],
 [4,5,6],
 [7,8,9]]

n = len(A)
for i in range(n+n-1):
 for j in range(i+1):
 k = i-j
 if k<n and k>=0 and j<n:
 print("對應(yīng)索引:",j,k, " 對應(yīng)值:",A[j][k])

A = [[1,2,3],
 [4,5,6],
 [7,8,9]]

# 得到下標(biāo)索引
size = len(A)
list_i = [] # 第一索引
list_j = [] # 第二索引正序
l1 = []  # 第二索引逆序
for m in range(2*size-1):
 for n in range(m+1):
 k = m-n
 if k<size and k>=0 and n<size:
  list_i.append(n)
  list_j.append(k)

print("第一索引i：\n", list_i)
print("第二索引j：")
for i in range(len(list_j)):
 a = list_j.pop()
 l1.append(a)
print(l1, "\n"*2, "輸出結(jié)果如下:")

for i in range(len(list_i)):
 print(A[list_i[i]][l1[i]], end= " ")

arr1 = [[1, 2, 3, 4],
 [5, 6, 7, 8],
 [9, 10, 11, 12],
 [13, 14, 15, 16]]

for each_arr in arr1:
 print(each_arr)

tem_arr = [] # 用來記錄數(shù)組值
rows = len(arr1)
cols = len(arr1[0])


def isValidIndex(x, n):
 return (x >= 0 and x < n)
 # 每一行的每個值的數(shù)組下標(biāo)的差都一樣，


for i in range(cols * 2 - 1): # 共輸出 cols * 2 - 1 行
 diff = cols - i - 1 # 每一行的差
 for j in range(cols): # 數(shù)組中每一個值的下標(biāo)范圍是0到cols
 k = j - diff # 通過一個下標(biāo)值計算另一個下標(biāo)值
 if isValidIndex(k, rows): # 剩下就是判斷這些下標(biāo)值是否滿足當(dāng)前的情況， 這一步不怎么好理解
  print(arr1[k][j], ' ', end='')
 print()