数组中局部极值的数量
原文： https://www.geeksforgeeks.org/maximum-number-local-extrema/
您将获得一个包含n个元素的数组。 极值是大于其两个邻居或小于其两个邻居的元素。 您必须计算给定数组中局部极值的数量。
注意：第一个和最后一个元素不是极值。
示例：
Input : a[] = {1, 5, 2, 5}
Output : 2
Input : a[] = {1, 2, 3}
Output : 0
方法：为了计算极值，我们必须检查一个元素是最大值还是最小值，即该元素是大于其两个邻居还是小于两个邻居。 为此，只需在数组上进行迭代，然后对每个元素检查其成为极值的可能性。
注意： a[0]和a[n-1]分别只有一个邻居，它们既不是最小值也不是最大值。
C++
// CPP to find number
// of extrema
#include <bits/stdc++.h>
using namespace std;
// function to find
// local extremum
int extrema(int a[], int n)
{
int count = 0;
// start loop from position 1
// till n-1
for (int i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
count += (a[i] > a[i - 1] && a[i] > a[i + 1]);
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
count += (a[i] < a[i - 1] && a[i] < a[i + 1]);
}
return count;
}
// driver program
int main()
{
int a[] = { 1, 0, 2, 1 };
int n = sizeof(a) / sizeof(a[0]);
cout << extrema(a, n);
return 0;
}
Java
// Java to find
// number of extrema
import java.io.*;
class GFG {
// function to find
// local extremum
static int extrema(int a[], int n)
{
int count = 0;
// start loop from
// position 1 till n-1
for (int i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
if(a[i] > a[i - 1] && a[i] > a[i + 1])
count += 1;
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
if(a[i] < a[i - 1] && a[i] < a[i + 1])
count += 1;
}
return count;
}
// driver program
public static void main(String args[])
throws IOException
{
int a[] = { 1, 0, 2, 1 };
int n = a.length;
System.out.println(extrema(a, n));
}
}
/* This code is contributed by Nikita Tiwari.*/
Python3
# Python 3 to find
# number of extrema
# function to find
# local extremum
def extrema(a, n):
count = 0
# start loop from
# position 1 till n-1
for i in range(1, n - 1) :
# only one condition
# will be true
# at a time either
# a[i] will be greater
# than neighbours or
# less than neighbours
# check if a[i] if
# greater than both its
# neighbours, then add
# 1 to x
count += (a[i] > a[i - 1] and a[i] > a[i + 1]);
# check if a[i] if
# less than both its
# neighbours, then
# add 1 to x
count += (a[i] < a[i - 1] and a[i] < a[i + 1]);
return count
# driver program
a = [1, 0, 2, 1 ]
n = len(a)
print(extrema(a, n))
# This code is contributed by Smitha Dinesh Semwal
C#
// C# to find
// number of extrema
using System;
class GFG {
// function to find
// local extremum
static int extrema(int []a, int n)
{
int count = 0;
// start loop from
// position 1 till n-1
for (int i = 1; i < n - 1; i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
if(a[i] > a[i - 1] && a[i] > a[i + 1])
count += 1;
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
if(a[i] < a[i - 1] && a[i] < a[i + 1])
count += 1;
}
return count;
}
// Driver program
public static void Main()
{
int []a = { 1, 0, 2, 1 };
int n = a.Length;
Console.WriteLine(extrema(a, n));
}
}
/* This code is contributed by vt_m.*/
PHP
<?php
// PHP to find number
// of extrema
// function to find
// local extremum
function extrema($a, $n)
{
$count = 0;
// start loop from position 1
// till n-1
for ($i = 1; $i < $n - 1; $i++)
{
// only one condition
// will be true at a
// time either a[i]
// will be greater than
// neighbours or less
// than neighbours
// check if a[i] is greater
// than both its neighbours
// then add 1 to x
$count += ($a[$i] > $a[$i - 1] and
$a[$i] > $a[$i + 1]);
// check if a[i] is
// less than both its
// neighbours, then
// add 1 to x
$count += ($a[$i] < $a[$i - 1] and
$a[$i] < $a[$i + 1]);
}
return $count;
}
// Driver Code
$a = array( 1, 0, 2, 1 );
$n = count($a);
echo extrema($a, $n);
// This code is contributed by anuj_67\.
?>
输出：
2
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