[LeetCode][C++] #1346. Check If N and Its Double Exist

[Easy][Question]:

Given an array arr of integers, check if there exist 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: For i = 0 and j = 2, arr[i] == 10 == 2 * 5 == 2 * arr[j]

Example 2:

Input: arr = [3,1,7,11]
Output: false
Explanation: There is no i and j that satisfy the conditions.

Constraints:

• 2 <= arr.length <= 500
• -103 <= arr[i] <= 103

My Solution[C++]:

[Ideas]: 這題第一直覺就是暴力解，先乘兩倍再開始一個一個找，但這樣時間複雜度是O(N^2)，所以要用另一個方式

利用HashTable的unordered_set來記錄，先做乘二或除二的判斷，是否存在Set裡面，如果有就結束，如果沒有就放進去Set（打不贏就加入的概念XD）