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 i
        		MAX = i;//最大的数 i
        		n += 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;
    }
};

8 ms 8.6 MB