给定一个
n×n的正整数列表，请编写一个函数，按照顺时针方向，从外围到中心重新排列元素。示例：
输入：[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
输出：[1, 2, 3, 6, 9, 8, 7, 4, 5]
题目难度：难
题目来源：codewars： Snail
def snail(nums: list) -> list:
pass
assert [[1,2,3],[4,5,6],[7,8,9]] == [1,2,3,6,9,8,7,4,5]
assert [[1,2,3],[8,9,4],[7,6,5]] == [1,2,3,4,5,6,7,8,9]
def snail(nums: list) -> list:
new_sort_list = []
while True:
new_sort_list += nums[0]
del nums[0]
if len(nums) == 0:
break
new_sort_list += [ns[len(ns) - 1] for ns in nums]
for ns in nums:
del ns[len(ns) - 1]
new_sort_list += nums[len(nums) - 1][::-1]
del nums[len(nums) - 1]
if len(nums) == 0:
break
new_sort_list += [ns[0] for ns in nums][::-1]
for ns in nums:
del ns[0]
if len(nums) == 0:
break
return new_sort_list
assert snail([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) == [1, 2, 3, 6, 9, 8, 7, 4, 5]
assert snail([[1, 2, 3], [8, 9, 4], [7, 6, 5]]) == [1, 2, 3, 4, 5, 6, 7, 8, 9]
assert snail([[1, 2, 3, 1], [4, 5, 6, 4], [7, 8, 9, 7], [7, 8, 9, 7]]) == [1, 2, 3, 1, 4, 7, 7, 9, 8, 7, 7, 4, 5, 6, 9,
8]
def test(dobleArr, list):
if len(dobleArr) > 0:
for i in dobleArr:
if dobleArr[0] == i:
for j in i:
list.append(j)
elif dobleArr[-1] == i:
for j in reversed(i):
list.append(j)
else:
list.append(i[-1])
i.remove(i[-1])
dobleArr.remove(dobleArr[0])
if len(dobleArr) > 0:
dobleArr.remove(dobleArr[-1])
for i in reversed(dobleArr):
list.append(i[0])
i.remove(i[0])
test(dobleArr, list)
return list
写复杂了
def get_list(nums_list: list):
# n数列
n = len(nums_list)
# 汇总数字
temp_list = []
for nums in nums_list:
for num in nums:
temp_list.append(num)
# 初始开始索引
x = 0
# 初始结束索引
y = n
# 定义初始幅度为正
is_plus = True
# 总共走的步数为2n-1，循环步数
times = 2 * n - 1
step_list = []
num_list = []
# 获取步调集合
for t in range(times):
# 第奇数次循环，步频为n，否则为1步
if t % 2 == 1:
step = n
else:
step = 1
# 判断是加还是减去
for a in range(2, times, 4):
if t == a:
is_plus = False
break
for b in range(0, times, 4):
if t == b:
is_plus = True
break
# 如果是减，步频为负数
if not is_plus:
step = -step
step_list.append(step)
s_index = 0
for s in step_list:
#每一步打印数字的初始索引
d = 0
for i in range(x, y, s):
num_list.append(temp_list[i])
d = d + 1
if s_index < len(step_list) - 1:
step_item_next = step_list[s_index + 1]
step_item = step_list[s_index]
#定义下一步骤集合的倍数，用于计算集合的结束值
if s_index % 2 == 0:
e = d - 2
else:
e = d - 1
if e < 0:
e = 0
if step_item < 0:
x = y + 1 + step_item_next
else:
x = y - 1 + step_item_next
if step_item_next > 0:
y = x + e * step_item_next + 1
else:
y = x + e * step_item_next - 1
s_index = s_index + 1
print(num_list)