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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]
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方法一
想法很简单，把list转成set 直接就去重了，然后从1到n，挨个判断是否在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 indexes
who points to a positive number.
we mark the number that i points as negative.
Then we filter the list, get all the indexes
who 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]) - 1nums[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]"""
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