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 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。
• 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[1,2,2,2,3,3,3,3,3]。
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 == 256
• 0 <= count[i] <= 10⁹
• 1 <= sum(count) <= 10⁹
• count 的众数是 唯一 的

思路

最小最大众数：遍历 平均：求和计数，注意溢出 中位数：二分

解法

class Solution {
    public double[] sampleStats(int[] count) {
        if (count == null || count.length != 256){
            return new double[]{0,0,0,0,0};
        }
        int i, j, cnt = 0, mid, target = 0;
        double[] res = new double[5];
        double sum = 0;
        res[0] = 256;
        res[1] = -1;
        long[] leftSum = new long[count.length];
        for (i = 0; i < 256; i++){
            if (res[0] == 256 && count[i] != 0 ){
                res[0] = i;
            }
            if ( count[i] != 0){
                res[1] = Math.max(res[1],i);
            }
            cnt += count[i];
            sum += count[i] * 1.0 * i;//count[i] * i会先溢出，再加给sum
            leftSum[i] = count[i];
            if (i > 0){
                leftSum[i] += leftSum[i - 1];
            }
            if (count[i] > count[((int)res[4])]){
                res[4] = i;
            }
        }
        res[2] = sum / cnt;
        
        target = (cnt + 1) / 2;
     
        i = 0;
        j = 255;
        while (i < j){
            mid = (i + j) / 2;
            if (leftSum[mid] >= target){
                j = mid;
            }else {
                i = mid + 1;
            }
        }
        if ((cnt & 1) == 0 && leftSum[i] == target){//偶数
            for (j = i + 1; j < 256; j++){
                if (count[j] != 0){
                    break;
                }
            }
            res[3] = (i + j) * 1.0 / 2;
        } else {
            res[3] = i;
        }
        
        
        return res;
    }
}

总结

• 分析出几种情况，然后分别对各个情况实现