代码拉取完成,页面将自动刷新
# Time: O(n * s * 2^n), s is the number of subset of which sum equals to side length.
# Space: O(n * (2^n + s))
class Solution(object):
def makesquare(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
total_len = sum(nums)
if total_len % 4:
return False
side_len = total_len / 4
fullset = (1 << len(nums)) - 1
used_subsets = []
valid_half_subsets = [0] * (1 << len(nums))
for subset in xrange(fullset+1):
subset_total_len = 0
for i in xrange(len(nums)):
if subset & (1 << i):
subset_total_len += nums[i]
if subset_total_len == side_len:
for used_subset in used_subsets:
if (used_subset & subset) == 0:
valid_half_subset = used_subset | subset
valid_half_subsets[valid_half_subset] = True
if valid_half_subsets[fullset ^ valid_half_subset]:
return True
used_subsets.append(subset)
return False
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。