Li Kou daily question - day 31 -1502 Judge whether an arithmetic sequence can be formed

2022.6.30 Did you brush the questions today ？

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subject ：

Here's an array of numbers arr .

If it's in a sequence , The difference between any two adjacent terms is always the same constant , Then this sequence is called A sequence of equal differences .

If you can rearrange the array to form an arithmetic sequence , Please return true ; otherwise , return false .

analysis ：

Give you an array , Determine whether the array elements form an arithmetic sequence , Be careful , Here the order is required ！

Ideas ： The reason why the summation formula cannot be used here is that some elements meet the summation but are not equal differences , So we use another property of isochronism

2*a（n2）=a（n1）+a（n3）

That's twice the median = Sum of previous and subsequent items

analysis ：

1. Violent solution

class Solution {
public:
    bool canMakeArithmeticProgression(vector<int>& arr) {
        sort(arr.begin(), arr.end());
        for (auto i = 1; i < arr.size()-1 ; i++)
        {
            if (2 * arr[i] != arr[i - 1] + arr[i + 1])
            {
                return false;
            }
        }
        return true;
    }
};