技術標籤：# 二分、位運算與數學java演算法leetcode

題目地址：

https://leetcode.com/problems/statistics-from-a-large-sample/

給定一個描述某個樣本的長 n n n陣列 A A A， A [ i ] A[i] A[i]代表 i i i出現的次數。問這個樣本的最小值、最大值、平均值、中位數和眾數。題目保證眾數一定存在。返回一個浮點型陣列。

最小值、最大值和眾數很容易求。要求平均值，要先求總和 ∑ i = 0 n − 1 i A [ i ] \sum_{i=0}^{n-1} iA[i] ∑i=0n−1​iA[i]和總個數 ∑ i = 0 n − 1 A [ i ] \sum_{i=0}^{n-1} A[i]

public class Solution {
    public double[] sampleStats(int[] count) {
        int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE, mode = 0, amount = 0, maxCount = 0;
		// 防止越界，要將sum定義為double型別
        double sum = 0;
        for (int i = 0; i < count.
 
length; i++) {
            if (count[i] > 0) {
                min = Math.min(min, i);
                max = Math.max(max, i);
                sum += i * count[i];
                amount += count[i];
                
                if (count[i] > maxCount) {
                    maxCount = count[i];
                    mode = i;
                }
            }
        }
        
        double average = sum / amount;
        int l = 0, r = 0;
        if (amount % 2 == 0) {
            l = amount / 2;
            r = l + 1;
        } else {
            l = r = amount / 2 + 1;
        }
        
        boolean odd = amount % 2 != 0;
        
        amount = 0;
        int posl = 0, posr = 0;
        boolean foundl = false, foundr = false;
        for (int i = 0; i < count.length; i++) {
            amount += count[i];
            if (amount >= r && !foundr) {
                posr = i;
                foundr = true;
            }
            if (amount >= l && !foundl) {
                posl = i;
                foundl = true;
            }
            
            if (foundl && foundr) {
                break;
            }
        }
        
        return new double[]{min, max, average, odd ? posl : (double) (posl + posr) / 2, mode};
    }
}

時間複雜度 O ( n ) O(n) O(n)，空間 O ( 1 ) O(1) O(1)。