Given an integer array arr, return the mean of the remaining integers after removing the smallest 5% and the largest 5% of the elements.
Answers within 10-5 of the actual answer will be considered accepted.
Example 1:
Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
Output: 2.00000
Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.
Example 2:
Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
Output: 4.00000
Example 3:
Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
Output: 4.77778
Example 4:
Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
Output: 5.27778
Example 5:
Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
Output: 5.29167
Constraints:
20 <= arr.length <= 1000
arr.length is a multiple of 20.
0 <= arr[i] <= 105
排序后去除首位计算平均数即可。
class MeanOfArrayAfterRemovingSomeElements : public Solution {
public:
void Exec() {
}
double trimMean(vector<int>& arr) {
std::sort(arr.begin(), arr.end());
int trimCount = arr.size() / 20;
double sum = 0;
for (int i = trimCount; i < arr.size() - trimCount; i++) {
sum += arr[i];
}
return sum / (arr.size() - 2 * trimCount);
}
};