比较直观好理解的解法：

class Solution {
public:
    int minOperations(int n) {
        int res = 0;
        for(int i = 1;i <= n;i++)
        {
            int num = 2*(i-1) + 1;
            if(num > n)
                res += num -n; 
        }
        
        return res;
    }
};

运行效率:

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Minimum Operations to Make Array Equal.
Memory Usage: 5.9 MB, less than 22.53% of C++ online submissions for Minimum Operations to Make Array Equal.

数列的平均值是n，如果当前数大于n,需要减去n即是需要操作的次数。