【LeetCode】448. Find All Numbers Disappeared in an Array【E】【77】

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本文介绍了一种高效查找数组中缺失数字的方法。对于长度为n的数组，其中包含1至n的整数，某些数字出现两次而其他数字仅出现一次。文章提供了一个不使用额外空间且运行时间为O(n)的解决方案，通过标记已出现的数字来快速找到未出现的数字。

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.

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]
        """