題目描述

我們對 0 到 255 之間的整數進行采樣，并將結果存儲在數組 count 中：count[k] 就是整數 k 的采樣個數。

我們以 浮點數 數組的形式，分別返回樣本的最小值、最大值、平均值、中位數和眾數。其中，眾數是保證唯一的。

我們先來回顧一下中位數的知識：

• 如果樣本中的元素有序，并且元素數量為奇數時，中位數為最中間的那個元素；
• 如果樣本中的元素有序，并且元素數量為偶數時，中位數為中間的兩個元素的平均值。

示例 1：

輸入：count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] 輸出：[1.00000,3.00000,2.37500,2.50000,3.00000]

示例 2：

輸入：count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] 輸出：[1.00000,4.00000,2.18182,2.00000,1.00000]

提示：

• count.length == 256
• 1 <= sum(count) <= 10^9
• 計數表示的眾數是唯一的
• 答案與真實值誤差在 10^-5 以內就會被視為正確答案

解法

直接求解即可，
求和防溢出，使用long類型

public double[] sampleStats(int[] count) {//最小值、最大值、平均值、中位數和眾數int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE, numCount=0, zs=0, countMax=0;long valCount=0;double[] res = new double[5];if(count == null || count.length<1) {return res;}for(int i=0;i<count.length;i++) {if(count[i]!=0) {min = Math.min(min, i);if(count[i] > countMax) {zs = i;}max = Math.max(max, i);countMax = Math.max(countMax, count[i]);numCount+= count[i];valCount+= i*count[i];}}if(numCount%2 ==0) {int tmp=0, first=0, second=0;for(int i=0;i<count.length;i++) {if(count[i]!=0) {tmp += count[i];if(tmp>=numCount/2 && first==0) {first = i;}if(tmp>=numCount/2+1 && second==0) {second=i;}if(first!=0 && second!=0) {res[3] = (double)(first+second)/(double)2;break;}}}}else {int tmp=0;for(int i=0;i<count.length;i++) {if(count[i]!=0) {tmp += count[i];if(tmp>=numCount/2) {res[3] = i;break;}}}}res[0] = min;res[1] = max;res[2] = (double)valCount/(double)numCount;res[4] = zs;return res;}

總結

以上是生活随笔為你收集整理的大样本统计的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。