找到最大峰值长度为 K 的子阵列
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，其中包含最大峰值。 线段【l，r】的峰是这样的指标，即 l < i < r 、a【I-1】<a【I】和a【I+1】<a【I】。 注:线段的边界指数 l 和 r 不是峰值。如果有许多子阵列具有最大峰值，则打印具有最小左索引的子阵列。 例:**
输入: arr = {3，1，4，1，5，9，2，6}，k = 7 输出: Left = 1 Right = 7 Peak = 2 说明: 有两个长度为 7 的子阵列，即【1，7】和【2，8】。两个子阵列都有 2 个峰值，即 3 和 6 指数是两个子阵列的峰值。我们必须返回具有最小 l 和最大峰值的子阵列，即 l = 1 和峰值= 2。 输入: arr = {3，2，3，2，1}，k = 3 输出: Left = 2 Right = 4 Peak = 1 解释: 只有一个长度为 3 且内部峰数为 1 的子阵列，即 l =2，Peak 为 i = 3。
方法: 解决这个问题的方法是使用一个滑动窗口，我们在 K 大小的窗口上滑动，找到每个窗口的峰值总数，哪个窗口给出的峰值数最大，哪个就是答案。当移动右索引时，我们检查添加的索引是否是峰值，我们增加计数，当移动左索引时，我们检查移除的索引是否是峰值，如果是，则减少计数。我们总是有一个 K 大小的窗户。 下面是上述方法的实现:
C++
// C++ implementation to Find subarray
// of Length K with Maximum Peak
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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 Singh
Python 3
# 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_Gangwar
C
// 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_ranjan
java 描述语言
<script>
// Javascript implementation to Find subarray
// of Length K with Maximum Peak
// Function to find the subarray
function findSubArray(a, n, k)
{
// Make prefix array to store
// the prefix sum of peak count
let pref = new Array(n);
pref[0] = 0;
for (let 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]++;
}
let peak = 0, left = 0;
for (let 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
document.write("Left = " + (left + 1) + "<br>");
document.write("Right = " + (left + k) + "<br>");
document.write("Peak = " + peak + "<br>");
}
// Driver code
let arr = [3, 2, 3, 2, 1];
let n = arr.length;
let k = 3;
findSubArray(arr, n, k);
// This code is contributed by gfgking
</script>
Output:
Left = 2
Right = 4
Peak = 1
时间复杂度: O(N)
版权属于：月萌API www.moonapi.com，转载请注明出处
