【leetcode】1346. Check If N and Its Double Exist

题目如下：

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

解题思路：方法很多，自由发挥吧。

代码如下：

class Solution(object):
    def checkIfExist(self, arr):
        """
        :type arr: List[int]
        :rtype: bool
        """
        dic = {}
        for i in arr:
            if i*2 in dic or (i%2 == 0 and i/2 in dic):
                return True
            dic[i] = 1
        return False

【leetcode】1346. Check If N and Its Double Exist

原文：https://www.cnblogs.com/seyjs/p/12287779.html