題目描述

我們對 0 到 255 之間的整數(shù)進(jìn)行采樣，并將結(jié)果存儲在數(shù)組 count 中：count[k] 就是整數(shù) k 的采樣個數(shù)。

我們以 浮點(diǎn)數(shù) 數(shù)組的形式，分別返回樣本的最小值、最大值、平均值、中位數(shù)和眾數(shù)。其中，眾數(shù)是保證唯一的。

我們先來回顧一下中位數(shù)的知識：

• 如果樣本中的元素有序，并且元素數(shù)量為奇數(shù)時，中位數(shù)為最中間的那個元素；
• 如果樣本中的元素有序，并且元素數(shù)量為偶數(shù)時，中位數(shù)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(shù)表示的眾數(shù)是唯一的
• 答案與真實值誤差在 10^-5 以內(nèi)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(shù)和眾數(shù)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;}

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