打印整数数组中的所有波峰和波谷
给定一个整数数组 arr[] ，任务是打印数组中所有波峰的列表和所有波谷的另一个列表。峰是数组中大于其相邻元素的元素。类似地，槽是比其相邻元素小的元素。
示例:
输入: arr[] = {5，10，5，7，4，3，5} 输出: 峰值:10 7 5 谷值:5 5 3 输入: arr[] = {1，2，3，4，5} 输出: 峰值:5 谷值:1
方法:对于数组的每个元素，检查当前元素是波峰(元素必须大于其相邻元素)还是波谷(元素必须小于其相邻元素)。 注意数组的第一个和最后一个元素将有一个邻居。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function that returns true if num is
// greater than both arr[i] and arr[j]
static bool isPeak(int arr[], int n, int num,
int i, int j)
{
// If num is smaller than the element
// on the left (if exists)
if (i >= 0 && arr[i] > num)
return false;
// If num is smaller than the element
// on the right (if exists)
if (j < n && arr[j] > num)
return false;
return true;
}
// Function that returns true if num is
// smaller than both arr[i] and arr[j]
static bool isTrough(int arr[], int n, int num,
int i, int j)
{
// If num is greater than the element
// on the left (if exists)
if (i >= 0 && arr[i] < num)
return false;
// If num is greater than the element
// on the right (if exists)
if (j < n && arr[j] < num)
return false;
return true;
}
void printPeaksTroughs(int arr[], int n)
{
cout << "Peaks : ";
// For every element
for (int i = 0; i < n; i++) {
// If the current element is a peak
if (isPeak(arr, n, arr[i], i - 1, i + 1))
cout << arr[i] << " ";
}
cout << endl;
cout << "Troughs : ";
// For every element
for (int i = 0; i < n; i++) {
// If the current element is a trough
if (isTrough(arr, n, arr[i], i - 1, i + 1))
cout << arr[i] << " ";
}
}
// Driver code
int main()
{
int arr[] = { 5, 10, 5, 7, 4, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
printPeaksTroughs(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true if num is
// greater than both arr[i] and arr[j]
static boolean isPeak(int arr[], int n, int num,
int i, int j)
{
// If num is smaller than the element
// on the left (if exists)
if (i >= 0 && arr[i] > num)
{
return false;
}
// If num is smaller than the element
// on the right (if exists)
if (j < n && arr[j] > num)
{
return false;
}
return true;
}
// Function that returns true if num is
// smaller than both arr[i] and arr[j]
static boolean isTrough(int arr[], int n, int num,
int i, int j)
{
// If num is greater than the element
// on the left (if exists)
if (i >= 0 && arr[i] < num)
{
return false;
}
// If num is greater than the element
// on the right (if exists)
if (j < n && arr[j] < num)
{
return false;
}
return true;
}
static void printPeaksTroughs(int arr[], int n)
{
System.out.print("Peaks : ");
// For every element
for (int i = 0; i < n; i++)
{
// If the current element is a peak
if (isPeak(arr, n, arr[i], i - 1, i + 1))
{
System.out.print(arr[i] + " ");
}
}
System.out.println("");
System.out.print("Troughs : ");
// For every element
for (int i = 0; i < n; i++)
{
// If the current element is a trough
if (isTrough(arr, n, arr[i], i - 1, i + 1))
{
System.out.print(arr[i] + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = {5, 10, 5, 7, 4, 3, 5};
int n = arr.length;
printPeaksTroughs(arr, n);
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
# Function that returns true if num is
# greater than both arr[i] and arr[j]
def isPeak(arr, n, num, i, j):
# If num is smaller than the element
# on the left (if exists)
if (i >= 0 and arr[i] > num):
return False
# If num is smaller than the element
# on the right (if exists)
if (j < n and arr[j] > num):
return False
return True
# Function that returns true if num is
# smaller than both arr[i] and arr[j]
def isTrough(arr, n, num, i, j):
# If num is greater than the element
# on the left (if exists)
if (i >= 0 and arr[i] < num):
return False
# If num is greater than the element
# on the right (if exists)
if (j < n and arr[j] < num):
return False
return True
def printPeaksTroughs(arr, n):
print("Peaks : ", end = "")
# For every element
for i in range(n):
# If the current element is a peak
if (isPeak(arr, n, arr[i], i - 1, i + 1)):
print(arr[i], end = " ")
print()
print("Troughs : ", end = "")
# For every element
for i in range(n):
# If the current element is a trough
if (isTrough(arr, n, arr[i], i - 1, i + 1)):
print(arr[i], end = " ")
# Driver code
arr = [5, 10, 5, 7, 4, 3, 5]
n = len(arr)
printPeaksTroughs(arr, n)
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if num is
// greater than both arr[i] and arr[j]
static Boolean isPeak(int []arr, int n, int num,
int i, int j)
{
// If num is smaller than the element
// on the left (if exists)
if (i >= 0 && arr[i] > num)
{
return false;
}
// If num is smaller than the element
// on the right (if exists)
if (j < n && arr[j] > num)
{
return false;
}
return true;
}
// Function that returns true if num is
// smaller than both arr[i] and arr[j]
static Boolean isTrough(int []arr, int n, int num,
int i, int j)
{
// If num is greater than the element
// on the left (if exists)
if (i >= 0 && arr[i] < num)
{
return false;
}
// If num is greater than the element
// on the right (if exists)
if (j < n && arr[j] < num)
{
return false;
}
return true;
}
static void printPeaksTroughs(int []arr, int n)
{
Console.Write("Peaks : ");
// For every element
for (int i = 0; i < n; i++)
{
// If the current element is a peak
if (isPeak(arr, n, arr[i], i - 1, i + 1))
{
Console.Write(arr[i] + " ");
}
}
Console.WriteLine("");
Console.Write("Troughs : ");
// For every element
for (int i = 0; i < n; i++)
{
// If the current element is a trough
if (isTrough(arr, n, arr[i], i - 1, i + 1))
{
Console.Write(arr[i] + " ");
}
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = {5, 10, 5, 7, 4, 3, 5};
int n = arr.Length;
printPeaksTroughs(arr, n);
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Function that returns true if num is
// greater than both arr[i] and arr[j]
function isPeak(arr, n, num, i, j)
{
// If num is smaller than the element
// on the left (if exists)
if (i >= 0 && arr[i] > num)
return false;
// If num is smaller than the element
// on the right (if exists)
if (j < n && arr[j] > num)
return false;
return true;
}
// Function that returns true if num is
// smaller than both arr[i] and arr[j]
function isTrough( arr, n, num, i, j)
{
// If num is greater than the element
// on the left (if exists)
if (i >= 0 && arr[i] < num)
return false;
// If num is greater than the element
// on the right (if exists)
if (j < n && arr[j] < num)
return false;
return true;
}
function printPeaksTroughs( arr, n)
{
document.write( "Peaks : ");
// For every element
for (var i = 0; i < n; i++) {
// If the current element is a peak
if (isPeak(arr, n, arr[i], i - 1, i + 1))
document.write( arr[i] + " ");
}
document.write( "<br>");
document.write( "Troughs : ");
// For every element
for (var i = 0; i < n; i++) {
// If the current element is a trough
if (isTrough(arr, n, arr[i], i - 1, i + 1))
document.write( arr[i] + " ");
}
}
var arr=[ 5, 10, 5, 7, 4, 3, 5 ];
printPeaksTroughs(arr, 7);
</script>
Output:
Peaks : 10 7 5
Troughs : 5 5 3
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