# leetcode1619.MeanofArrayA

### 描述

``````Given an integer array arr, return the mean of the remaining integers after removing the smallest 5% and the largest 5% of the elements.

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

Note:

20 <= arr.length <= 1000
arr.length is a multiple of 20.
0 <= arr[i] <= 10^5

### 解答

``````class Solution(object):
def trimMean(self, arr):
"""
:type arr: List[int]
:rtype: float
"""
arr.sort()
N = int(len(arr) * 0.05)
result = arr[N:-N]
return float(sum(result))/len(result)

### 运行结果

``````Runtime: 40 ms, faster than 77.37% of Python online submissions for Mean of Array After Removing Some Elements.
Memory Usage: 13.5 MB, less than 71.53% of Python online submissions for Mean of Array After Removing Some Elements.

### 解答

``````class Solution(object):
def trimMean(self, arr):
"""
:type arr: List[int]
:rtype: float
"""
arr.sort()
for _ in range(int(len(arr) * .05)):
arr.pop(0)
arr.pop()
return float(sum(arr)) / len(arr)

### 运行结果

``````Runtime: 32 ms, faster than 99.15% of Python online submissions for Mean of Array After Removing Some Elements.
Memory Usage: 13.8 MB, less than 19.66% of Python online submissions for Mean of Array After Removing Some Elements.