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

class Solution {
    public boolean checkIfExist(int[] arr) {
        Map<Integer, Integer> map = new HashMap();
        int zero = 0;
        for(int i = 0; i < arr.length; i++){
            map.putIfAbsent(arr[i], i);
            if(arr[i] == 0) zero++;
        }
        for(int i: arr){
            if(zero == 2) return true;
            else{
                map.remove(0);
                if(map.containsKey(2*i)) return true;
            }
        }
        return false;
    }
}

0很特殊，如果用hashmap的话要注意判断0的数量，如果是两个就是true，一个的话要先remove了0再做判断。

class Solution {
    public boolean checkIfExist(int[] arr) {
        HashSet<Integer> set = new HashSet<>();
        for(int a : arr) {
            if(set.contains(a*2) || (a%2 == 0 && set.contains(a/2))) return true;
            set.add(a);
        }
        return false;
    }
}

相比而言hashset更好