📅  最后修改于: 2023-12-03 15:32:46.401000             🧑  作者: Mango
在一个包含m个元素的数组中找到两个子集,使得这两个子集的差异最大,并返回这个差异值。
首先,我们需要找到所有可能的子集。可以使用递归的方式来实现这个过程。接下来,我们需要计算每个子集的差异,并找到具有最大差异的两个子集。最后,返回它们之间的差异值。
def find_max_difference(nums):
subsets = get_subsets(nums)
max_diff = None
for i in range(len(subsets)):
for j in range(i+1, len(subsets)):
diff = abs(sum(subsets[i]) - sum(subsets[j]))
if max_diff is None or diff > max_diff:
max_diff = diff
return max_diff
def get_subsets(nums, index=0):
if index == len(nums):
return [[]]
else:
subsets = get_subsets(nums, index+1)
item = nums[index]
new_subsets = [subset + [item] for subset in subsets]
return subsets + new_subsets
nums = [1, 2, 3]
find_max_difference(nums) # Output: 2
在这个例子中,可能的子集为:[[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]。其中,[3]和[1, 2]这两个子集之间的差异为2,这也是最大的差异值。