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[LeetCode] Subsets

时间:2015-06-03 00:50:27      阅读:263      评论:0      收藏:0      [点我收藏+]

Well, this problem becomes fairly easy once we use the itertools package of Python. It is just a transformation of the type of the result. And other things are already done by itertools.

The idea is to create a list res. Then generate subsets of size 0, 1, ..., len(nums) of numsusing the combinations method of itertools, and transform them to list (originally tuple) and then append them to res.

Of course, this is not a meaningful solution and this problem does not aim at testing whether we know how to use itertools. The key idea still lies at backtracking.

 1 class Solution:
 2     # @param {integer[]} nums
 3     # @return {integer[][]}
 4     def subsets(self, nums):
 5         nums.sort() 
 6         res = []
 7         for sz in range(len(nums) + 1):
 8             combs = itertools.combinations(nums, sz)
 9             for comb in combs:
10                 res.append(list(comb))
11         return res 

The following solutions are more desirable.

 1 class Solution:
 2     # @param {integer[]} nums
 3     # @return {integer[][]}
 4     def subsets(self, nums):
 5         nums.sort() 
 6         res = []
 7         res.append([])
 8         for num in nums:
 9             n = len(res)
10             for idx in range(n):
11                 temp = list(res[idx])
12                 temp.append(num)  
13                 res.append(temp)
14         return res
15 
16 class Solution:
17     # @param {integer[]} nums
18     # @return {integer[][]}
19     def subsets(self, nums):
20         nums.sort()
21         num_subset = 2 ** len(nums)
22         res = [[]] * num_subset
23         for i in range(len(nums)):
24             for j in range(num_subset):
25                 if j >> i & 1:
26                     temp = list(res[j]) 
27                     temp.append(nums[i])
28                     res[j] = list(temp)
29         return res

 

[LeetCode] Subsets

原文:http://www.cnblogs.com/jcliBlogger/p/4548112.html

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