Article directory
Topic link and description
https://leetcode.cn/problems/mean-of-array-after-removing-some-elements/Give
you an integer array arr, please delete the minimum 5% of the numbers and the maximum 5% of the numbers, the remaining numbers average value.
Results within 10-5 of the standard answer were considered correct.
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 removing the largest and smallest elements in the array, all elements are equal to 2, so the average 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
hint:
20 <= arr.length <= 1000
arr.length is a multiple of 20
0 <= arr[i] <= 105
Key words:
After unpretentious sorting, take the average of the middle 90%
method one:
run screenshot
the code
public double trimMean(int[] arr) {
Arrays.sort(arr);
// 题目说了是20的倍数，所以不担心浮点数
int range = arr.length / 20;
// 这里需要考虑一下精度问题
double total = 0;
for (int i = range; i < arr.length - range; i++) {
total += arr[i];
}
return total / (arr.length - (range << 1));
}
end
Welcome to communicate in the comment area, check in every day, and rush! ! !