leetcode No448. Find All Numbers Disappeared in an Array

Posted 2022-06-11 Dufre.WC

Question

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.

Algorithm&Code

第一时间想到的是hash了，不过只beats不到20%的submit，在Discuss中看到一个很快的解法，我先贴hash的。

class Solution 
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) 
        int n = nums.size();
        unordered_map<int, int> hash;
        vector<int> res;
        
        for(int i=0;i<n;i++)
            hash[nums[i]]++;
        
        for(int i=1;i<=n;i++)
            if(hash[i] == 0)
                res.push_back(i);
        
        
        return res;
    
;

现在来看快一点的解法，其实要注意到题目中说到some elements appear twice and others appear once.
我也没太看懂，不过看到一个评论就懂了。
e.g. [ 4, 3, 2, 7, 8, 2, 3, 1] we can use a binary array (O(N) space) to count it. [ 1, 1, 1, 1, 0, 0, 1, 1]
Note that it has the same amount of information with [ -, -, -, -, +, +, -, -] . see?

c++

class Solution 
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) 
        for(int i=0;i<nums.size();i++)
            int index = abs(nums[i])-1;
            nums[index] = nums[index] > 0? -nums[index]: nums[index];
        
        
        vector<int> res;
        for(int i=0;i<nums.size();i++)
            if(nums[i]>0)
                res.push_back(i+1);
        
        
        return res;
    
;

python：

class Solution:
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        for i in nums:
            index = abs(i) - 1
            nums[index] = -abs(nums[index])
            
        res = []
        for i in range(0, len(nums)):
            if nums[i] > 0:
                res.append(i+1)
                
        return res