LeetCode-1093 大样本统计

题目链接：

源自LeetCode第142场周赛

https://leetcode-cn.com/contest/weekly-contest-142/problems/statistics-from-a-large-sample/

题目描述：

我们对 0 到 255 之间的整数进行采样，并将结果存储在数组 count 中：count[k] 就是整数 k 的采样个数。

我们以 浮点数 数组的形式，分别返回样本的最小值、最大值、平均值、中位数和众数。其中，众数是保证唯一的。

我们先来回顾一下中位数的知识：

• 如果样本中的元素有序，并且元素数量为奇数时，中位数为最中间的那个元素；
• 如果样本中的元素有序，并且元素数量为偶数时，中位数为中间的两个元素的平均值。

示例 1：

输入：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]
输出：[1.00000,3.00000,2.37500,2.50000,3.00000]

示例 2：

输入：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]
输出：[1.00000,4.00000,2.18182,2.00000,1.00000]

提示：

1. count.length == 256
2. 1 <= sum(count) <= 10^9
3. 计数表示的众数是唯一的
4. 答案与真实值误差在 10^-5 以内就会被视为正确答案

题目大意：

给出一个数组count[256]，count[i]表示数字i（i大于等于0小于等于255）在一个序列中出现的次数，要求统计出序列的最小值、最大值、平均数、中位数和众数。

思路：

最小值就是第一个count[i]大于0的i；最大值就是最后一个count[i]>0的i；平均数是对于所有count[i]>0，i*count[i]的加和除以序列中数字的个数cnt；找到最大的count[i]，i就是众数。

中位数要分情况讨论。定义一个新变量nextcount=0，类似于cnt的统计，重新遍历count数组，对于每个count[i]>0，nextcnt为出现i之前所有出现的数的个数。

若统计出来的序列中数字的个数cnt为奇数，则nextcnt<=cnt/2并且nextcnt+count[i]>cnt/2时，此时i就是中位数。例如cnt=5，序列为{1,2,2,3,5}，cnt/2=5/2=2，i为2时，nextcnt=1，count[i]=2，nextcnt=1<2并且nextcnt+count[2]=3>2，2就是中位数。

若统计出来的序列中数字的个数cnt为偶数，则要对nextcnt分情况讨论，如果nextcnt已经加到了cnt/2（即nextcnt==cnt/2），则上一个出现的数pre一定是偶数个数字的序列中运算得到中位数所需的第一部分，当前遍历到的count[i]>0的i就是运算得到中位数所需的第二部分，(第一部分+第二部分)/2即得到中位数，例如cnt=6，序列为{1,2,2,4,4,5}，i为4时，pre=2，nextcnt=3=cnt/2，nextcnt+count[4]=3+2=5>cnt/2，中位数=(pre+i)/2=(2+4)/2=3；还有一种是nextcntcnt/2，此时i就是中位数，例如cnt=6，序列为{1,3,3,3,3,5}，i为3时，nextcnt=1cnt/2，3就是中位数。

上AC代码：

class Solution {
public:
    vector sampleStats(vector& count) {
        vector ret;
        int i,j;
        //众数
        double most=-1;
        //中位数
        double medium=-1;
        //最小值
        double xiao=0x3f3f3f3f;
        //最大值
        double da=-0x3f3f3f3f;
        //平均值
        double pingjun=-1;
        //出现次数最多的数出现了多少次
        int maxsum=-0x3f3f3f3f;
        //用于统计序列中数的个数
        int cnt=0;
        //序列和 用于求平均值
        double sum=0;
        
        for(i=0;i<256;i++)
        {
            if(count[i]>0)
            {
                //找最大
                if(i>da)
                {
                    da=i;
                }
                //找最小
                if(imaxsum)
                {
                    most=i;
                    maxsum=count[i];
                }
            }
        }
        
        ret.push_back(xiao);
        ret.push_back(da);
        //求平均值
        pingjun=sum/cnt;
        ret.push_back(pingjun);
        //找中位数
        int nextcnt=0;
        if(cnt%2==1)
        {
            //奇数
            for(i=0;i<256;i++)
            {
                if(count[i]>0)
                {
                    if(nextcnt<=cnt/2&&nextcnt+count[i]>cnt/2)
                    {
                        medium=i;
                        break;
                    }
                    nextcnt+=count[i];
                }
            }
        }
        else
        {
            //偶数
            int pre=-1;
            for(i=0;i<256;i++)
            {
                if(count[i]>0)
                {
                    if(nextcntcnt/2)
                    {
                        medium=i;
                        break;
                    }
                    else if(nextcnt==cnt/2&&nextcnt+count[i]>cnt/2)
                    {
                        medium=(i+pre)/2.0;
                        break;
                    }
                    nextcnt+=count[i];
                    pre=i;
                }
            }
        }
        ret.push_back(medium);
        ret.push_back(most);
        
        return ret;
    }
};

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