1. Find the minimum rotation of the array
Description rotating array, such as [-2, 3, 5, 6, 9, -10, -5], is characterized by the sequence of the left array to make a good part moves to the right.
[Thinking] binary search
def find_min(li):
l = 0
r = len(li)
m = int(r/2)
if li[l] < li[m]:
# 此时是左边递增，最小值位于右边
return find_min(li[m+1:r])
elif li[l] > li[m]:
# 最小值位于mid左面，包含mid
return find_min(li[l:m+1])
else:
return li[mid]2. Zero and arrays
Description element and the sub-array closest to 0
[idea] referred to 1-n and the elements s (n), then there are two cases, S (n) and S (m) -S (n) . The second of which can be sorted, it does take two neighboring poor. Sort complexity nlog (n).
from collections import OrderedDict
from copy import deepcopy
def get_mini_sublist(li):
part1_index = {}
part2_index = {}
# 第一部分,前n项求和
for i in range(1, len(li)):
part1_index[(0, i)] = sum(li[:i])
part1_index = OrderedDict(sorted(part1_index.items(), key=lambda x:x[1] ))
# 第二部分，排序，做差
t1 = deepcopy(part1_index).popitem(last=True),
t2 = deepcopy(part1_index).popitem(last=False)
for i,j in zip(t1.items(), t2.items()):
part2_index[(i[0][1], j[0][1])] = j[1]-i[1]
# 融合，然后取值最小值
part2_index.update(part1_index)
key_min = min(part2_index, key=lambda x:abs(part2_index[x])
return key_min, part2_index[key_min]