摘要：前言的大樣本統計我們對到之間的整數進行采樣，并將結果存儲在數組中就是整數的采樣個數。我們以浮點數數組的形式，分別返回樣本的最小值最大值平均值中位數和眾數。

Weekly Contest 142的 大樣本統計：

我們對 0 到 255 之間的整數進行采樣，并將結果存儲在數組 count 中：count[k] 就是整數 k 的采樣個數。
我們以 浮點數 數組的形式，分別返回樣本的最小值、最大值、平均值、中位數和眾數。其中，眾數是保證唯一的。
我們先來回顧一下中位數的知識：

如果樣本中的元素有序，并且元素數量為奇數時，中位數為最中間的那個元素；
如果樣本中的元素有序，并且元素數量為偶數時，中位數為中間的兩個元素的平均值。

示例1：
輸入：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]
輸出：[1.00000,3.00000,2.37500,2.50000,3.00000]
示例2：
輸入：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]
輸出：[1.00000,4.00000,2.18182,2.00000,1.00000]
提示：

count.length == 256
1 <= sum(count) <= 10^9
計數表示的眾數是唯一的
答案與真實值誤差在 10^-5 以內就會被視為正確答案

本地難度為中等，首先需要讀懂題目意思，本題的入參數組count其實算是一個壓縮數據后的數組。

我們對 0 到 255 之間的整數進行采樣，并將結果存儲在數組 count 中：count[k] 就是整數 k 的采樣個數。

簡單來說就是，數組count的第k個元素就是k在壓縮前的數組中出現count[k]個。以示例1的count為例

[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]

解壓的過程如下：

第0個元素為0，則解壓后的數組為[]
第1個元素為4，則解壓后的數組為[1,1,1,1]
第2個元素為3，則解壓后的數組為[1,1,1,1,2,2,2]
第3個元素為2，則解壓后的數組為[1,1,1,1,2,2,2,3,3]
第4個元素為2，則解壓后的數組為[1,1,1,1,2,2,2,3,3,4,4]
......
省略后續步驟

搞清楚count的數據特征后，選擇使用TreeMap對count進行處理，將有效數字及其出現個數存儲起來（有效數字指的是count[k]不為0的元素）。根據就是根據題目要求分別處理以下指標：

最小值：TreeMap中第一個key

最大值：TreeMap中最后一個key

平均值：TreeMap的key之和除以value之和

中位數：

計算出數組實際的元素個數（即value之和）

根據元素個數的奇偶性，獲取對應的值

眾數：出現次數最多的數字，即TreeMap中value最大的鍵值對的key

    /**
     * 1093. 大樣本統計
     *
     * @param count
     * @return
     */
    public double[] sampleStats(int[] count) {
        // 使用TreeMap有序存儲數字及其出現次數
        TreeMap countMap = new TreeMap<>();
        double[] result = new double[5];
        // 總和
        double sum = 0L;
        // 數字出現總次數
        double total = 0L;
        // 最大出現次數
        long maxTimes = 0;
        // 最小值
        double min;
        // 最大值
        double max;
        // 平均值
        double average;
        // 中位數
        double middle = 0;
        // 眾數，出現次數最多的數字
        double mode = 0;
        for (int i = 0; i < count.length; i++) {
            if (count[i] != 0) {
                countMap.put(i, count[i]);
                sum = sum + i * count[i];
                total += count[i];
                if (count[i] > maxTimes) {
                    maxTimes = count[i];
                    mode = i;
                }
            }
        }
        min = countMap.firstKey().doubleValue();
        max = countMap.lastKey().doubleValue();
        average = sum / total;
        // 是否為奇數
        boolean odd = total % 2 != 0;
        // 中位數索引
        int middleIndex = (int) ((total - 1) / 2);
        int index = -1;
        Iterator> it = countMap.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry entry = it.next();
            int num = entry.getKey();
            int times = entry.getValue();
            index += times;
            if (index > middleIndex) {
                middle = num;
                break;
            } else if (index == middleIndex) {
                if (odd) {
                    middle = num;
                    break;
                } else {
                    middle = (num + it.next().getKey()) / 2.0;
                    break;
                }
            }
        }
        result[0] = min;
        result[1] = max;
        result[2] = average;
        result[3] = middle;
        result[4] = mode;
        return result;
    }