Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

思路：

取一个新list，里面元素从1到n，然后通过set相减得到没有在已知数组中出现的元素，然后转换为list。
建议：使用xrange，xrange比range快

class Solution(object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        news = xrange(1, len(nums)+1)
        #return [i for i in news if i not in list(set(nums))]
        return list(set(news) - set(nums))