[LeetCode]448. Find All Numbers

题目

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]

难度

Easy

方法

因为1 ≤ a[i] ≤ n,将a[abs(a[i])-1]置为负数，最后遍历数组a，如果a[i]>0，则i+1未在数组a中出现

python代码

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]