Leetcode 448. Find All Numbers Disappeared in an Array
这是道Leetcode easy题，但比较有意思的点是，题目要求用O(n) 的时间复杂度和O(1)的空间复杂度。要同时满足这个条件，还是需要一点技巧的。 具体题目如下：
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：需要两个循环：第一个循环，遍历原列表nums，对该数的数值在nums的位置-1处，该该位置的值设为负值（表示该位置+1这个数曾经出现过），第二个循环，遍历更改后的nums，大于0的数所在的位置+1即为未出现的数，将这些数输出即可。逻辑很简单，但比较难解释清楚，具体看代码：

class Solution:
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        for i in range(len(nums)):
            m = abs(nums[i])-1
            if nums[m]>0: 
                nums[m] = -nums[m]
        output = []
        for i in range(len(nums)):
            if nums[i]>0:
                output.append(i+1)
        return output