LeetCode 1093. 大样本统计

1. 题目

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/statistics-from-a-large-sample
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

class Solution {
public:vector<double> sampleStats(vector<int>& count) {int i, MIN = -1, MAX, avg, median = 0, mode;int n = 0, ni = 0, N;long sum = 0;for(i = 0; i < count.size(); ++i){if(count[i]){if(MIN == -1)MIN = i;//第一个最小的数 iMAX = i;//最大的数 in += count[i];//个数累加sum += count[i]*i;//求和if(count[i] > ni)//单个数 最大的个数{ni = count[i];mode = i;//众数}}}vector<double> ans({MIN,MAX,double(sum)/n,0,mode});ni = 0;bool flag = false;for(i = 0; i < count.size(); ++i){	//寻找中位数if(n%2)//总的个数是奇数个{if(ni+count[i] < n/2+1)ni += count[i];else{ans[3] = i;break;}}else//偶数个{N = ni + count[i];if(N < n/2)ni += count[i];else if(N >= n/2 && N < n/2+1){ans[3] = i;flag = true;//中位数，卡在两个数上}else{if(!flag){ans[3] = i;break;}else{ans[3] = double(ans[3]+i)/2;break;}}}}return ans;}
};