LeetCode-Python-1093. 大样本统计

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：

1. count.length == 256
2. 1 <= sum(count) <= 10^9
3. 计数表示的众数是唯一的
4. 答案与真实值误差在 10^-5 以内就会被视为正确答案

思路：

首先要搞清输入的格式：

[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]

代表原数组[1, 2, 2, 2, 3, 3, 3, 3]。count[i] = k 就代表有 k 个数值为 i 的数

然后题目需要求最小值、最大值、平均值、中位数和众数。

最小值很简单，count数组从左往右第一个不是0的数的下标就是最小值。

最大值同理，从右往左在count数组里找第一个不是0的数的下标。

平均值就需要先统计原数组里数字的个数 = sum（count），然后计算原数组里所有元素之和 = sum（count[i] * i），除一下即可。

中位数需要找到恰好排在中间的一个数或者两个数，线性扫描count，把每个元素的值用一个变量累加，

如果累加和>总个数的一半，说明中位数已经出现了，

然后再判断一下是只有一个数还是有两个数。

众数就是找到count数组里最大值对应的下标。

class Solution(object):
    def sampleStats(self, count):
        """
        :type count: List[int]
        :rtype: List[float]
        """
        s = 0
        total_cnt = sum(count)
        cnt  = 0
        avg, median, mean, mean_cnt = 0, 0, 0, 0
        min_element, max_element = 0, 0
        #找最小值
        for i in range(len(count)):
            if count[i] != 0:
                min_element = i
                break
        #找最大值
        for i in range(len(count) - 1, -1, -1):
            if count[i] != 0:
                max_element = i
                break
                
        #找一共统计了多少个数字        
        geshu = 0
        for i in count:
            if i > 0:
                geshu += i
                
        find = False
        for i, num in enumerate(count):
            s += num * i #计算总和
            if mean_cnt < num: #找count数组最大值的下标
                mean = i
                mean_cnt = num
                
            cnt += num #找目前出现了多少个数字
            if cnt > total_cnt // 2 and find == False: 
                if total_cnt % 2: #中位数肯定是一个数
                    median = i
                    find = True
                else:
                    if cnt - num == total_cnt // 2: #中位数有两个不同的数
                        for j in range(i - 1, -1, -1): #往前找上一个数
                            if count[j] > 0:
                                median = (i + j) /2.0
                                find = True
                                break
                    else:#中位数有两个相同的数
                        median = i
                        find = True
                                
        return [min_element, max_element, 1.0 * s /geshu, median, mean ]