find-all-numbers-disappeared-in-an-array.py

from __future__ import print_function
# Time:  O(n)
# Space: O(1)

# 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]


class Solution(object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        for i in xrange(len(nums)):
            if nums[abs(nums[i]) - 1] > 0:
                nums[abs(nums[i]) - 1] *= -1

        result = []
        for i in xrange(len(nums)):
            if nums[i] > 0:
                result.append(i+1)
            else:
                nums[i] *= -1
        return result

    def findDisappearedNumbers2(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        return list(set(range(1, len(nums) + 1)) - set(nums))

    def findDisappearedNumbers3(self, nums):
        for i in range(len(nums)):
            index = abs(nums[i]) - 1
            nums[index] = - abs(nums[index])

        return [i + 1 for i in range(len(nums)) if nums[i] > 0]


if __name__ == '__main__':
    s = Solution()
    r = s.findDisappearedNumbers([4, 3, 2, 7, 8, 2, 3, 1])
    print(r)