Open carloscn opened 1 year ago
pub fn trim_mean(arr: Vec<i32>) -> f64
{
let mut ret:f64 = 0.0f64;
if arr.len() < 1 {
return ret;
}
let mut arr_dup:Vec<i32> = arr.clone();
let mut cut_len:usize = (arr.len() as f32 * 0.05_f32) as usize;
let mut sum:i32 = 0;
arr_dup.sort();
for i in cut_len..arr.len() - cut_len {
sum += arr_dup[i];
}
ret = (sum as f64) / (arr.len() - 2 * cut_len) as f64;
return ret;
}
问题描述
给你一个整数数组 arr ,请你删除最小 5% 的数字和最大 5% 的数字后,剩余数字的平均值。
与 标准答案 误差在 10-5 的结果都被视为正确结果。
示例 1:
输入:arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3] 输出:2.00000 解释:删除数组中最大和最小的元素后,所有元素都等于 2,所以平均值为 2 。
示例 2:
输入:arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0] 输出:4.00000
示例 3:
输入: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] 输出:4.77778
示例 4:
输入: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] 输出:5.27778
示例 5:
输入: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] 输出:5.29167
提示:
20 <= arr.length <= 1000 arr.length 是 20 的 倍数 0 <= arr[i] <= 105
来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/mean-of-array-after-removing-some-elements