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]
Subscribe to see which companies asked this question.方法一想法很簡單，把list轉成set 直接就去重了，然后從1到n，挨個判斷是否在set里方法二For each number i in nums,we mark the number that i points as negative.Then we filter the list, get all the indexeswho points to a positive number.Since those indexes are not visited.
class Solution(object):    def findDisappearedNumbers(self, nums):        for i in xrange(len(nums)):            ind = abs(nums[i]) - 1            nums[ind] = - abs(nums[ind])        res = []        for i in xrange(len(nums)):            if nums[i] > 0:                res += i+1,        return res        '''        l = len(nums)        nums = set(nums)        res = []        for i in range(1,l+1):            if i not in nums:                res += i,        return res        '''        """        :type nums: List[int]        :rtype: List[int]        """