给定一个长度为 n 的数组 arr[]和一个正整数 K ，我们必须找到一个长度为 K 的子数组，它里面有最大的峰值。
段 [l, r] 的峰值是那些满足l < i < r 、 a[i-1] < a[i]和a[i+1] < a[i] 的索引。
注意：该段的边界索引 l 和 r 不是峰值。如果有许多具有最大峰值的子数组，则打印具有最小左索引的子数组。
例子：
Input :
arr = {3, 1, 4, 1, 5, 9, 2, 6}, k = 7
Output:
Left = 1
Right = 7
Peak = 2
Explanation:
There are two subarray with length 7 i.e [1, 7] and [2, 8]. Both subarray has 2 peak inside it i.e 3 and 6 index are the peak in both the subarray. We have to return the subarray with minimum l and maximum peak i.e l = 1 and peak = 2.
Input:
arr = {3, 2, 3, 2, 1}, k = 3
Output :
Left = 2
Right = 4
Peak = 1
Explanation:
Only one subarray whose length is 3 and number of peak inside it is 1 i.e. l =2 and peak is i = 3.
方法：
解决这个问题的方法是使用滑动窗口，我们在 K 的窗口大小上滑动，并找到每个窗口中的峰值总数，给出最大峰值数的窗口将是答案。在移动右侧索引时，我们检查添加的索引是否为峰值，我们增加计数，而在移动左侧索引时，我们检查删除的索引是否为峰值，如果是，则减少计数。我们总是有一个大小为 K 的窗口。
下面是上述方法的实现：
C++
// C++ implementation to Find subarray
// of Length K with Maximum Peak
#include
using namespace std;
// Function to find the subarray
void findSubArray(int* a, int n, int k)
{
// Make prefix array to store
// the prefix sum of peak count
int pref[n];
pref[0] = 0;
for (int i = 1; i < n - 1; ++i) {
// Count peak for previous index
pref[i] = pref[i - 1];
// Check if this element is a peak
if (a[i] > a[i - 1] && a[i] > a[i + 1])
// Increment the count
pref[i]++;
}
int peak = 0, left = 0;
for (int i = 0; i + k - 1 < n; ++i)
// Check if number of peak in the sub array
// whose l = i is greater or not
if (pref[i + k - 2] - pref[i] > peak) {
peak = pref[i + k - 2] - pref[i];
left = i;
}
// Print the result
cout << "Left = " << left + 1 << endl;
cout << "Right = " << left + k << endl;
cout << "Peak = " << peak << endl;
}
// Driver code
int main()
{
int arr[] = { 3, 2, 3, 2, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 3;
findSubArray(arr, n, k);
return 0;
} Java
// Java implementation to Find subarray
// of Length K with Maximum Peak
class GFG{
// Function to find the subarray
static void findSubArray(int []a, int n, int k)
{
// Make prefix array to store
// the prefix sum of peak count
int []pref = new int[n];
pref[0] = 0;
for (int i = 1; i < n - 1; ++i) {
// Count peak for previous index
pref[i] = pref[i - 1];
// Check if this element is a peak
if (a[i] > a[i - 1] && a[i] > a[i + 1])
// Increment the count
pref[i]++;
}
int peak = 0, left = 0;
for (int i = 0; i + k - 1 < n; ++i)
// Check if number of peak in the sub array
// whose l = i is greater or not
if (pref[i + k - 2] - pref[i] > peak) {
peak = pref[i + k - 2] - pref[i];
left = i;
}
// Print the result
System.out.print("Left = " + (left + 1) +"\n");
System.out.print("Right = " + (left + k) +"\n");
System.out.print("Peak = " + peak +"\n");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3, 2, 3, 2, 1 };
int n = arr.length;
int k = 3;
findSubArray(arr, n, k);
}
}
// This code contributed by Princi SinghPython3
# Python3 implementation to Find subarray
# of Length K with Maximum Peak
# Function to find the subarray
def findSubArray(a, n, k):
# Make prefix array to store
# the prefix sum of peak count
pref = [0 for i in range(n)]
pref[0] = 0
for i in range(1, n - 1, 1):
# Count peak for previous index
pref[i] = pref[i - 1]
# Check if this element is a peak
if (a[i] > a[i - 1] and a[i] > a[i + 1]):
# Increment the count
pref[i] += 1
peak = 0
left = 0
for i in range(0, n - k + 1, 1):
# Check if number of peak in the sub array
# whose l = i is greater or not
if (pref[i + k - 2] - pref[i] > peak):
peak = pref[i + k - 2] - pref[i]
left = i
# Print the result
print("Left =",left + 1)
print("Right =",left + k)
print("Peak =",peak)
# Driver code
if __name__ == '__main__':
arr = [3, 2, 3, 2, 1]
n = len(arr)
k = 3
findSubArray(arr, n, k)
# This code is contributed by Surendra_GangwarC#
// C# implementation to Find subarray
// of Length K with Maximum Peak
using System;
class GFG{
// Function to find the subarray
static void findSubArray(int []a, int n, int k)
{
// Make prefix array to store
// the prefix sum of peak count
int []pref = new int[n];
pref[0] = 0;
for(int i = 1; i < n - 1; ++i)
{
// Count peak for previous index
pref[i] = pref[i - 1];
// Check if this element is a peak
if (a[i] > a[i - 1] && a[i] > a[i + 1])
{
// Increment the count
pref[i]++;
}
}
int peak = 0;
int left = 0;
for(int i = 0; i + k - 1 < n; ++i)
{
// Check if number of peak in the sub array
// whose l = i is greater or not
if (pref[i + k - 2] - pref[i] > peak)
{
peak = pref[i + k - 2] - pref[i];
left = i;
}
}
// Print the result
Console.Write("Left = " + (left + 1) + "\n");
Console.Write("Right = " + (left + k) + "\n");
Console.Write("Peak = " + peak + "\n");
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 3, 2, 3, 2, 1 };
int n = arr.Length;
int k = 3;
findSubArray(arr, n, k);
}
}
// This code is contributed by Rohit_ranjanJavascript
输出：
Left = 2
Right = 4
Peak = 1时间复杂度： O(N)
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