数组中局部极值的数量 · GeeksForGeeks 中文系列教程

# 数组中局部极值的数量 > 原文： [https://www.geeksforgeeks.org/maximum-number-local-extrema/](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 // 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 // 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 ```py # 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# ```cs // 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 // 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 ``` * * * * * *