原題：

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,If nums = [1,2,3], a solution is:

[  [3],  [1],  [2],  [1,2,3],  [1,3],  [2,3],  [1,2],  []]這道題就是生成子集的問題，高中知識，子集個數為2^n個。所以我的思路是0-2^n的循環，然后將十進制數對應到二進制，根據二進制的值對應到某個元素是否在某個子集中。詳細代碼（代碼不好，速度有點慢，leetcode上有大神的代碼，估計C語言會快一點？）:
class Solution(object):    def subsets(self, nums):        """        :type nums: List[int]        :rtype: List[List[int]]        """        length = len(nums)  # size of nums        result = []         if length == 0:            return result        for x in xrange(0, 2**length):            s = bin(x)            sub = []            n = x            # decimal to binary            for i in xrange(0, length):                if n == 0:                    break                else:                    if n % 2 == 1:                        sub.append(nums[i])                    n = n / 2            result.append(sub)        return result
其中十進制to二進制的代碼，求余求商，求出的結果是從低位到高位（對應學的十進制轉化二進制的方式，結果從下到上排列），上邊用的時候有些許變化：
n = 12PRint bin(n)while n != 0:    print n%2    n = n/2其實python中有十進制到二進制轉化的函數: bin()
bin(x)Convert an integer number to a binary string. The result is a valid Python expression. If x is not a Python int object, it has to define an __index__() method that returns an integer.
轉化為了字符串，10 - 0b1010，注意0b也是字符串的內容，所以真正的二進制字符串是從索引2開始的。