原文:
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
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
output:
peaks : 10 7 5
troughs : 5 5 3
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