数组中范围平均值的Java程序

给定一个包含 n 个整数的数组。给你 q 个查询。编写一个程序，在新行中为每个查询打印从 l 到 r 范围内的均值的下限值。

例子 ：

Input : arr[] = {1, 2, 3, 4, 5}
        q = 3
        0 2
        1 3
        0 4
Output : 2
         3
         3
Here for 0 to 2 (1 + 2 + 3) / 3 = 2

Input : arr[] = {6, 7, 8, 10}
        q = 2
        0 3
        1 2
Output : 7
         7

天真的方法：我们可以为每个查询 l 到 r 运行循环，并找到范围内元素的总和和数量。在此之后，我们可以打印每个查询的均值下限。

Java

// Java program to find floor value
// of mean in range l to r
public class Main {
  
    // To find mean of range in l to r
    static int findMean(int arr[], int l, int r)
    {
        // Both sum and count are
        // initialize to 0
        int sum = 0, count = 0;
  
        // To calculate sum and number
        // of elements in range l to r
        for (int i = l; i <= r; i++) {
            sum += arr[i];
            count++;
        }
  
        // Calculate floor value of mean
        int mean = (int)Math.floor(sum / count);
  
        // Returns mean of array
        // in range l to r
        return mean;
    }
  
    // Driver program to test findMean()
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5 };
        System.out.println(findMean(arr, 0, 2));
        System.out.println(findMean(arr, 1, 3));
        System.out.println(findMean(arr, 0, 4));
    }
}

输出 ：

2
3
3

时间复杂度： O(n)

有效方法：我们可以使用前缀求和的数字来找到数字的总和。 prefixSum[i] 表示前 i 个元素的总和。因此，从 l 到 r 范围内的数字总和将是 prefixSum[r] – prefixSum[l-1]。 l 到 r 范围内的元素数将为 r – l + 1。所以我们现在可以在 O(1) 中打印范围 l 到 r 的平均值。

Java

// Java program to find floor value
// of mean in range l to r
public class Main {
public static final int MAX = 1000005;
    static int prefixSum[] = new int[MAX];
  
    // To calculate prefixSum of array
    static void calculatePrefixSum(int arr[], int n)
    {
        // Calculate prefix sum of array
        prefixSum[0] = arr[0];
        for (int i = 1; i < n; i++)
            prefixSum[i] = prefixSum[i - 1] + arr[i];
    }
  
    // To return floor of mean
    // in range l to r
    static int findMean(int l, int r)
    {
        if (l == 0)
           return (int)Math.floor(prefixSum[r] / (r + 1));
          
        // Sum of elements in range l to
        // r is prefixSum[r] - prefixSum[l-1]
        // Number of elements in range
        // l to r is r - l + 1
        return (int)Math.floor((prefixSum[r] -
                prefixSum[l - 1]) / (r - l + 1));
    }
  
    // Driver program to test above functions
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5 };
        int n = arr.length;
        calculatePrefixSum(arr, n);
        System.out.println(findMean(1, 2));
        System.out.println(findMean(1, 3));
        System.out.println(findMean(1, 4));
    }
}

输出：

2
3
3

有关更多详细信息，请参阅有关数组范围平均值的完整文章！