3194. Minimum Average of Smallest and Largest Elements

題目敘述:

You have an array of floating point numbers averages which is initially empty. You are given an array nums of n integers where n is even.

You repeat the following procedure n / 2 times:

Remove the smallest element, minElement, and the largest element maxElement, from nums.
Add (minElement + maxElement) / 2 to averages.
Return the minimum element in averages.

class Solution {
public:
    double minimumAverage(vector<int>& v) {
        int n=v.size();
        sort(v.begin(),v.end());
        int i=0,j=n-1;
        double ans=1000;
        while(i<j)
        {
            double k=(v[i]+v[j])/(2*1.0);
            ans=min(ans,k);
            i++;
            j--;
        }
        return ans;
    }
};