1093. 大样本统计

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

算法：

class Solution {
public:
    vector sampleStats(vector& count) {
        int size= count.size();
        double mina = 256.0;
        double maxa = 0.0;

        vector>res;
        for(int i=0;i=0; i--)
        {
            if(count[i]!=0)
            {
                maxa = double(i);
                break;
            }
        }
        double sum = 0.0;
        long long c = 0;
        for(int i=0;i c/2)
            {
                zhongweishu = double(v.first);
                break;
            }
                
            else
            {
                if(c%2==1)
                {
                    flag1 = true;
                }
                else
                {
                    flag1=true;
                    k1 = v.first;
                }
            }      
        }
        double zhongshu = 0.0;
        int maxv = -1;
        
        for(int i=0;im;
        
        m.push_back(mina);
        m.push_back(maxa);
         m.push_back(pinjun);
         m.push_back(zhongweishu);
         m.push_back(zhongshu);
        return m;
    }
};