1346. Check If N and Its Double Exist

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

这题不难，但是排序后判断有一个陷阱，负数要特别处理一下。

class CheckIfNAndItsDoubleExists : public Solution {
public:
    void Exec() {
        vector<int> input{-10,12,-20,-8,15};
        auto res = checkIfExist(input);
    }
    bool checkIfExist(vector<int>& arr) {
        std::sort(arr.begin(), arr.end());

        for (int i = 0; i < arr.size() - 1; i++) {
            for (int j = i + 1; j < arr.size(); j++) {
                if (arr[i] * 2 == arr[j] || arr[j] * 2 == arr[i]) {
                    return true;
                }
            }
        }
        return false;
    }
};