寻找平均值最小的子阵列的 Java 程序
原文:https://www . geesforgeks . org/Java-program-for-find-the-subarray-with-min-average/
给定大小为 n 且整数为 k 的数组 arr[,使得 k <= n。
示例:
Input: arr[] = {3, 7, 90, 20, 10, 50, 40}, k = 3
Output: Subarray between indexes 3 and 5
The subarray {20, 10, 50} has the least average
among all subarrays of size 3.
Input: arr[] = {3, 7, 5, 20, -10, 0, 12}, k = 2
Output: Subarray between [4, 5] has minimum average
我们强烈建议您点击此处进行练习，然后再进入解决方案。
一个简单的解决方案是把每个元素看作大小为 k 的子阵列的开始，并从这个元素开始计算子阵列的和。这个解决方案的时间复杂度是 O(nk)。
一个高效的解决方案就是在 O(n)个时间和 O(1)个额外空间内解决上述问题。想法是使用大小为 k 的滑动窗口。跟踪当前 k 个元素的总和。要计算当前窗口的总和，请移除前一个窗口的第一个元素并添加当前元素(当前窗口的最后一个元素)。
1) Initialize res_index = 0 // Beginning of result index
2) Find sum of first k elements. Let this sum be 'curr_sum'
3) Initialize min_sum = sum
4) Iterate from (k+1)'th to n'th element, do following
for every element arr[i]
a) curr_sum = curr_sum + arr[i] - arr[i-k]
b) If curr_sum < min_sum
res_index = (i-k+1)
5) Print res_index and res_index+k-1 as beginning and ending
indexes of resultant subarray.
下面是上述算法的实现。
Java 语言(一种计算机语言，尤用于创建网站)
// A Simple Java program to find
// minimum average subarray
class Test {
static int arr[] = new int[] { 3, 7, 90, 20, 10, 50, 40 };
// Prints beginning and ending indexes of subarray
// of size k with minimum average
static void findMinAvgSubarray(int n, int k)
{
// k must be smaller than or equal to n
if (n < k)
return;
// Initialize beginning index of result
int res_index = 0;
// Compute sum of first subarray of size k
int curr_sum = 0;
for (int i = 0; i < k; i++)
curr_sum += arr[i];
// Initialize minimum sum as current sum
int min_sum = curr_sum;
// Traverse from (k+1)'th element to n'th element
for (int i = k; i < n; i++)
{
// Add current item and remove first
// item of previous subarray
curr_sum += arr[i] - arr[i - k];
// Update result if needed
if (curr_sum < min_sum) {
min_sum = curr_sum;
res_index = (i - k + 1);
}
}
System.out.println("Subarray between [" +
res_index + ", " + (res_index + k - 1) +
"] has minimum average");
}
// Driver method to test the above function
public static void main(String[] args)
{
int k = 3; // Subarray size
findMinAvgSubarray(arr.length, k);
}
}
输出:
Subarray between [3, 5] has minimum average
时间复杂度:O(n) 辅助空间:O(1)
详情请参考整篇文章找到平均值最小的子阵！
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