一、大样本统计（Weekly Contest 142）

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：

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

思路：
题目的意思就是count[index] = k,表示出现了k个index，这样就明白了
中位值的计算可能需要一些技巧

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
class Solution {
    public double[] sampleStats(int[] count) {
    int max = Integer.MIN_VALUE;
    	int min = Integer.MAX_VALUE;
    	double sum = 0;
    	double []result = new double[5];
    	int index = 0;
    	int size = 0;
    	int zhongshu = 0;
    	int time = 0;
    	for (int i = 0; i < count.length; i++) {
			if(count[i] != 0){
				min = i;
				break;
			}
		}
    	Map<Integer, Integer> getMap = new TreeMap<>();
    	for (int i = count.length - 1; i >= 0; i--) {
    		if(count[i] != 0){
				max = i;
				break;
			}
		}
    	for (int d : count) {
    		if(time < d){
    			time = d;
    			zhongshu = index;
    		}
    		sum += d * index;
    		if(d != 0){
    			getMap.put(index, d);
    		}
    		index ++;
    		size += d;
		}
    	
    	result[0] = min;
    	result[1] = max;
    	result[2] = (double)(sum / (size * 1.0));
    	int mid = 0;
    	int num1 = 0;
    	int num2 = 0;
    	boolean flag = true;
    	for (Entry<Integer, Integer> integer : getMap.entrySet()) {
			mid += integer.getValue();
			if(mid >= size / 2  && flag){
				num1 = integer.getKey();
				flag = false;
			}
			if(mid >= size / 2 + 1){
				num2 = integer.getKey();
				break;
			}
    	}
  
    	if(size % 2 == 1){
    		result[3] = num1;
    	}else {
    		result[3] = (num1 + num2)  / 2.0;
		}
    	result[4] = zhongshu;
    	return result;
    }
}

代码写的比较复杂了，取最大和最小值我是分开了

这个代码就显得比较简单了

class Solution {
    public double[] sampleStats(int[] count) {
        double[] result = new double[5];
        int minV = -1,maxV=-1,ct = 0,maxC=-1;
        long sum = 0;
        for(int i=0;i<count.length;++i){
            if(count[i]!=0){
                maxV=i;
                if(minV==-1)
                    minV=i;
                if(maxC==-1||count[i]>count[maxC])
                    maxC=i;
            }
            sum+=i*count[i];
            ct+=count[i];
        }
        int findCt = (ct+1)/2;
        int temp = 0;
        int ctV = 0;
        for(int i=0;i<count.length;++i){
            if(count[i]>0) {
                findCt -= count[i];
                if (findCt <= 0) {
                    ctV += i;
                    break;
                }
            }
        }
        findCt = (ct+1)/2;
        for(int i=count.length-1;i>=0;--i){
            if(count[i]>0) {
                findCt -= count[i];
                if (findCt <= 0) {
                    ctV += i;
                    break;
                }
            }
        }
        result[0] = (double)minV;
        result[1] = (double)maxV;
        result[2] = (double)sum;
        result[3] = (double)ctV;
        result[4] = (double) maxC;
        result[2]/=ct;
        result[3]/=2;
        return result;

    }
}