題目

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]

難度

Easy

方法

因為1 ≤ a[i] ≤ n,將a[abs(a[i])-1]置為負數，最后遍歷數組a，如果a[i]>0，則i+1未在數組a中出現

python代碼

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

        result = []
        for i in range(len(nums)):
            if nums[i] > 0:
                result.append(i+1)

        return result

assert Solution().findDisappearedNumbers([4,3,2,7,8,2,3,1]) == [5,6]