'''
Given an integer array, find a subarray where the sum of numbers is zero.
Your code should return the index of the first number and the index of the last number.
Example
Given [-3, 1, 2, -3, 4], return [0, 2] or [1, 3].
Note
There is at least one subarray that it's sum equals to zero.
'''
'''
# 1 两个for循环穷举
class Solution:
def call(self, arr, sum_obj_elem, once=True):
size = len(arr)
result = []
for i1 in range(size):
tp = sum_obj_elem
for i2 in range(i1, size):
tp -= arr[i2]
if tp == 0:
print(i1, i2)
result.append([i1, i2])
if once:
return result
return result
s = Solution()
print(s.call([-3, 1, 2, -3, 4], 0, False))
'''
# 分奇偶对讨论，奇：按中轴划分-3, 1, 2。偶：按左右划分1, 6, 2, -1, -2, -6
# 一句话总结：xxxxx子串和为0，只存在奇/偶两个个数情况，所有列举两种情况时从中间展开计算和是否为零即可（奇：1 1 1, 2, -1 -1 -3;偶:1 2 , -1 -2）
class Solution2:
def call(self, arr, sum_obj_elem, once=True):
size = len(arr)
result = []
# 计算奇数形式
for center in range(size):
if once and len(result) > 0:
break
left_len = center
right_len = size - 1 - center
# print(left_len, right_len)
sum = 0
left = center
right = center
if left_len < right_len:
d = left_len
else:
d = right_len
sum += arr[center]
while d > 0 and left - 1 > 0 and right + 1 < size:
d -= 1
left -= 1
right += 1
sum += arr[left]
sum += arr[right]
if sum == 0:
result.append(arr[left:right + 1])
break
# 计算双数形式1,1,-1,-1
for center in range(size - 1):
if once and len(result) > 0:
break
left_len = center + 1
right_len = size - 1 - center
d = min(left_len, right_len)
left = center
right = center + 1
# print([left], [right])
sum = 0
while d > 0 and left >= 0 and right < size:
d -= 1
sum += arr[left]
sum += arr[right]
if sum == 0:
result.append(arr[left:right + 1])
break
left -= 1
right += 1
return result
s = Solution2()
print(s.call([5, -3, 1, 2, -3, 4, 1, 6, 2, -1, -2, -6], 0, False))
#[[-3, 1, 2], [1, 2, -3], [1, 6, 2, -1, -2, -6]]
