"Leetcode" Find all Numbers disappeared the problem-solving report in an Array

"Leetcode" Find all Numbers disappeared the problem-solving report in an Array

[Leetcode]

Https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/Total accepted:14302 Total Submissions: 24993 difficulty:easy Question

Given an array of integers where 1≤a[i]≤n (n = size of array),
Some elements appear twice and others appear.

Find all the elements of [1, N] inclusive which does not appear in this
Array.

Could do it without extra spaces and in O (n) runtime? May
Assume the returned list does not count as extra spaces. Example:

Input: [4,3,2,7,8,2,3,1]

Output: [5,6] Ways

At first I did not think of a very effective method, can only use violence to solve. A single method that does not meet the topic without additional space requirements

Method One:

Class Solution (object):
    def finddisappearednumbers (self, Nums): ""
        : Type Nums:list[int]
        : rtype: List[int] ""
        "
        cList = List (range (1, len (nums) + 1))
        returnlist = [] for
        x in Nums:
            clist[x-1] = 0
  for x in CList:
            if x!= 0:
                returnlist.append (x) return
        returnlist

ac:359 ms

Method Two:

Referring to other people's, I learned a way: in situ negative to mark. For example, for [4, 3, 2, 7, 8, 2, 3, 1], these elements as the index of the list, points to the element to transform into a negative, then, there is no change to negative position is no one pointed to it, so this position corresponding subscript did not appear.

Class Solution (object):
    def finddisappearednumbers (self, Nums): ""
        : Type Nums:list[int]
        : Rtype:list[int] "" For
        I in range (len (nums)):
            index=abs (Nums[i])-1
            nums[index]=-ABS (nums[ Index]) return
        [i+1 to I in range (len (nums)) if nums[i] > 0]

ac:362 ms

This speed is still not ideal. Date

January 2, 2017