448. Find All Numbers Disappeared in an Array@python

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]

原题地址: Find All Numbers Disappeared in an Array

难度: Easy

题意: 给出n个数,范围在[1, n]之间,数组中存在重复,返回1-n之中缺少的数,要求不用额外的空间,时间复杂度为O(n)

不用额外的空间,只能用数组进行计数

(1)遍历数组,用索引进行计数,出现一次将索引值变为负数.数值[1, n]对应的索引为[0, n-1]比如上个例子中4,索引值为3,所以将数值7变为-7

(2)遍历一次之后,重复一次的值的索引对应的值为正数,而其他数都为负数, 再遍历一次,找出值为正数的索引值

代码:

class Solution(object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        n = len(nums)
        res = []
        for i in range(n):
            idx = abs(nums[i]) - 1    # 索引值
            nums[idx] = -abs(nums[idx]) 

        for i in range(n):
            if nums[i] > 0:
                res.append(i+1)
        return res

时间复杂度:O(n)

空间复杂度:O(1)