目录链接：
力扣编程题-解法汇总_分享+记录-CSDN博客
GitHub同步刷题项目：
https://github.com/September26/java-algorithms
原题链接：力扣
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 在样本中出现的次数。
计算以下统计数据:
minimum：样本中的最小元素。maximum：样品中的最大元素。mean：样本的平均值，计算为所有元素的总和除以元素总数。median：- 如果样本的元素个数是奇数，那么一旦样本排序后，中位数
median就是中间的元素。 - 如果样本中有偶数个元素，那么中位数
median就是样本排序后中间两个元素的平均值。
- 如果样本的元素个数是奇数，那么一旦样本排序后，中位数
mode：样本中出现次数最多的数字。保众数是 唯一 的。
以浮点数数组的形式返回样本的统计信息 [minimum, maximum, mean, median, mode] 。与真实答案误差在 10-5 内的答案都可以通过。
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] 解释：用count表示的样本为[1,2,2,2,3,3,3,3]。 最小值和最大值分别为1和3。 均值是(1+2+2+2+3+3+3+3) / 8 = 19 / 8 = 2.375。 因为样本的大小是偶数，所以中位数是中间两个元素2和3的平均值，也就是2.5。 众数为3，因为它在样本中出现的次数最多。
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] 解释：用count表示的样本为[1,1,1,1,2,2,3,3,3,4,4]。 最小值为1，最大值为4。 平均数是(1+1+1+1+2+2+2+3+3+4+4)/ 11 = 24 / 11 = 2.18181818…(为了显示，输出显示了整数2.18182)。 因为样本的大小是奇数，所以中值是中间元素2。 众数为1，因为它在样本中出现的次数最多。
提示：
count.length == 2560 <= count[i] <= 1091 <= sum(count) <= 109-
count的众数是 唯一 的
解题思路：
/**
* 最小，最大，平均值，最多都比较简单，重点是中位数。
* 因为数量很多，我们挨个去统计肯定不合适，
* 所以我们首先可以总数量，然后分别求matchleft和matchright代表左右的中位数，如果是基数的话matchright=matchright。
* 然后我们不断的sum去减掉count[i]的值，当num <= matchright时，说明已经走到了右侧的中位数，这时候记录其值到medianValue。num <= matchleft时也是一样。
* 最后medianValue/2就是我们的预期中位数。
*/
代码：
#include <iostream>
#include <map>
#include <list>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
/**
* 最小，最大，平均值，最多都比较简单，重点是中位数。
* 因为数量很多，我们挨个去统计肯定不合适，
* 所以我们首先可以总数量，然后分别求matchleft和matchright代表左右的中位数，如果是基数的话matchright=matchright。
* 然后我们不断的sum去减掉count[i]的值，当num <= matchright时，说明已经走到了右侧的中位数，这时候记录其值到medianValue。num <= matchleft时也是一样。
* 最后medianValue/2就是我们的预期中位数。
*/
class Solution
{
public:
vector<double> sampleStats(vector<int> &count)
{
vector<double> list;
double minValue = -1;
double maxValue = -1;
double meanValue = 0;
double medianValue = 0;
double modeValue = 0;
// 最大数量
int maxNum = 0;
// 总数量
int num = 0;
// 总的累加值
double sum = 0;
for (int i = 0; i < count.size(); i++)
{
if (count[i] == 0)
{
continue;
}
num += count[i];
sum += (count[i] * (long)i);
maxValue = i;
if (minValue == -1)
{
minValue = i;
}
if (count[i] > maxNum)
{
modeValue = i;
maxNum = count[i];
}
}
meanValue = sum / num;
// 求中位数
int matchleft = 0;
int matchright = 0;
if (num % 2 == 0)
{
matchleft = num / 2 - 1;
matchright = num / 2;
}
else
{
matchleft = num / 2;
matchright = num / 2;
}
for (int i = 0; i < count.size(); i++)
{
num -= count[i];
if (num <= matchright)
{
medianValue += i;
matchright = -1;
}
if (num <= matchleft)
{
medianValue += i;
break;
}
}
medianValue = medianValue / 2;
list.push_back(minValue);
list.push_back(maxValue);
list.push_back(meanValue);
list.push_back(medianValue);
list.push_back(modeValue);
return list;
}
};
