1346. Check If N and Its Double Exist (Easy)

Q

Given an array arr of integers, check if there exists two integers N and M such that N is the double of M ( i.e. N = 2 * M).

More formally check if there exists two indices i and j such that :

• i != j
• 0 <= i, j < arr.length
• arr[i] == 2 * arr[j]

Example 1:

Input: arr = [10,2,5,3]
Output: true
Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.

Example 2:

Input: arr = [7,1,14,11]
Output: true
Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.

Example 3:

Input: arr = [3,1,7,11]
Output: false
Explanation: In this case does not exist N and M, such that N = 2 * M.

Constraints:

• 2 <= arr.length <= 500
• -10^3 <= arr[i] <= 10^3

A:
忘记 0 的特殊情况了。

class Solution {
public:
    bool checkIfExist(vector<int>& arr) {
        std::set<int> s;
        int count0 = 0;
        for(int a : arr)
        {
            if (a==0)            
                count0 +=1;
            else
                s.insert(a);
        }
        if (count0>=2)
            return true;
        for(int a:arr)
        {
            if(s.find(a+a) != s.end())
            {
                return true;
            }
        }
        return false;
    }
};