1. 題目

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

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/statistics-from-a-large-sample
著作權歸領扣網絡所有。商業轉載請聯系官方授權，非商業轉載請注明出處。

2. 解題

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;//第一個最小的數 iMAX = i;//最大的數 in += count[i];//個數累加sum += count[i]*i;//求和if(count[i] > ni)//單個數 最大的個數{ni = count[i];mode = i;//眾數}}}vector<double> ans({MIN,MAX,double(sum)/n,0,mode});ni = 0;bool flag = false;for(i = 0; i < count.size(); ++i){ //尋找中位數if(n%2)//總的個數是奇數個{if(ni+count[i] < n/2+1)ni += count[i];else{ans[3] = i;break;}}else//偶數個{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;//中位數，卡在兩個數上}else{if(!flag){ans[3] = i;break;}else{ans[3] = double(ans[3]+i)/2;break;}}}}return ans;} };

8 ms 8.6 MB

總結

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