给定N个整数的圆形数组arr [] ,以使给定数组的最后一个元素与数组的第一个元素相邻,任务是在此圆形数组中打印Next Greater Element。对于不存在更大元素的元素,请将下一个更大元素视为“ -1” 。
例子:
Input: arr[] = {5, 6, 7}
Output: {6, 7, -1}
Explanation:
Next Greater Element for 5 is 6, for 6 is 7, and for 7 is -1 as we don’t have any element greater than itself so its -1.
Input: arr[] = {4, -2, 5, 8}
Output: {5, 5, 8, -1}
Explanation:
Next Greater Element for 4 is 5, for -2 its 5, for 5 is 8, and for 8 is -1 as we don’t have any element greater than itself so its -1, and for 3 its 4.
方法:此问题可以使用贪婪方法解决。步骤如下:
- 为了使圆形数组的属性有效,请再次将给定的数组元素附加到同一数组。
例如:
Let arr[] = {1, 4, 3}
After appending the same set of elements arr[] becomes
arr[] = {1, 4, 3, 1, 4, 3}
- 找到下一个更大的元素,直到上述数组中的N个元素形成为止。
- 如果找到更大的元素,则打印该元素,否则打印“ -1” 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the NGE
void printNGE(int A[], int n)
{
// Formation of cicular array
int arr[2 * n];
// Append the given array element twice
for (int i = 0; i < 2 * n; i++)
arr[i] = A[i % n];
int next, i, j;
// Iterate for all the
// elements of the array
for (i = 0; i < n; i++) {
// Initialise NGE as -1
next = -1;
for (j = i + 1; j < 2 * n; j++) {
// Checking for next
// greater element
if (arr[i] < arr[j]) {
next = arr[j];
break;
}
}
// Print the updated NGE
cout << next << ", ";
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 2, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
printNGE(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the NGE
static void printNGE(int []A, int n)
{
// Formation of cicular array
int []arr = new int[2 * n];
// Append the given array element twice
for(int i = 0; i < 2 * n; i++)
arr[i] = A[i % n];
int next;
// Iterate for all the
// elements of the array
for(int i = 0; i < n; i++)
{
// Initialise NGE as -1
next = -1;
for(int j = i + 1; j < 2 * n; j++)
{
// Checking for next
// greater element
if (arr[i] < arr[j])
{
next = arr[j];
break;
}
}
// Print the updated NGE
System.out.print(next + ", ");
}
}
// Driver Code
public static void main(String args[])
{
// Given array arr[]
int []arr = { 1, 2, 1 };
int N = arr.length;
// Function call
printNGE(arr, N);
}
}
// This code is contributed by Code_Mech
Python3
# Python3 program for the above approach
# Function to find the NGE
def printNGE(A, n):
# Formation of cicular array
arr = [0] * (2 * n)
# Append the given array
# element twice
for i in range(2 * n):
arr[i] = A[i % n]
# Iterate for all the
# elements of the array
for i in range(n):
# Initialise NGE as -1
next = -1
for j in range(i + 1, 2 * n):
# Checking for next
# greater element
if(arr[i] < arr[j]):
next = arr[j]
break
# Print the updated NGE
print(next, end = ", ")
# Driver code
if __name__ == '__main__':
# Given array arr[]
arr = [ 1, 2, 1 ]
N = len(arr)
# Function call
printNGE(arr, N)
# This code is contributed by Shivam Singh
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the NGE
static void printNGE(int []A, int n)
{
// Formation of cicular array
int []arr = new int[2 * n];
// Append the given array element twice
for(int i = 0; i < 2 * n; i++)
arr[i] = A[i % n];
int next;
// Iterate for all the
// elements of the array
for(int i = 0; i < n; i++)
{
// Initialise NGE as -1
next = -1;
for(int j = i + 1; j < 2 * n; j++)
{
// Checking for next
// greater element
if (arr[i] < arr[j])
{
next = arr[j];
break;
}
}
// Print the updated NGE
Console.Write(next + ", ");
}
}
// Driver Code
public static void Main()
{
// Given array arr[]
int []arr = { 1, 2, 1 };
int N = arr.Length;
// Function call
printNGE(arr, N);
}
}
// This code is contributed by Code_Mech
Javascript
C++
// C++ program to demonstrate the use of circular
// array without using extra memory space
#include
using namespace std;
// Function to find the Next Greater Element(NGE)
void printNGE(int A[], int n)
{
int k;
for (int i = 0; i < n; i++) {
// Initialise k as -1 which is printed when no NGE
// found
k = -1; //
for (int j = i + 1; j < n + i; j++) {
if (A[i] < A[j % n]) {
printf("%d ", A[j % n]);
k = 1;
break;
}
}
if (k == -1) // Gets executed when no NGE found
printf("-1 ");
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 8, 6, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
printNGE(arr, N);
return 0;
}
Java
// Java program to demonstrate the
// use of circular array without
// using extra memory space
import java.io.*;
class GFG{
// Function to find the Next
// Greater Element(NGE)
static void printNGE(int A[], int n)
{
int k;
for(int i = 0; i < n; i++)
{
// Initialise k as -1 which is
// printed when no NGE found
k = -1;
for(int j = i + 1; j < n + i; j++)
{
if (A[i] < A[j % n])
{
System.out.print(A[j % n] + " ");
k = 1;
break;
}
}
// Gets executed when no NGE found
if (k == -1)
System.out.print("-1 ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int[] arr = { 8, 6, 7 };
int N = arr.length;
// Function call
printNGE(arr, N);
}
}
// This code is contributed by yashbeersingh42
Python3
# Python3 program to demonstrate the use of circular
# array without using extra memory space
# Function to find the Next Greater Element(NGE)
def printNGE(A, n) :
for i in range(n) :
# Initialise k as -1 which is printed when no NGE
# found
k = -1
for j in range(i + 1, n + i) :
if (A[i] < A[j % n]) :
print(A[j % n], end = " ")
k = 1
break
if (k == -1) : # Gets executed when no NGE found
print("-1 ", end = "")
# Given array arr[]
arr = [ 8, 6, 7 ]
N = len(arr)
# Function call
printNGE(arr, N)
# This code is contributed by divyeshrabadia07
C#
// C# program to demonstrate the
// use of circular array without
// using extra memory space
using System;
class GFG {
// Function to find the Next
// Greater Element(NGE)
static void printNGE(int[] A, int n)
{
int k;
for(int i = 0; i < n; i++)
{
// Initialise k as -1 which is
// printed when no NGE found
k = -1;
for(int j = i + 1; j < n + i; j++)
{
if (A[i] < A[j % n])
{
Console.Write(A[j % n] + " ");
k = 1;
break;
}
}
// Gets executed when no NGE found
if (k == -1)
Console.Write("-1 ");
}
}
static void Main()
{
// Given array arr[]
int[] arr = { 8, 6, 7 };
int N = arr.Length;
// Function call
printNGE(arr, N);
}
}
// This code is contributed by divyesh072019
2, -1, 2,
此方法需要O(n 2 )时间,但要占用O(n)阶的额外空间
一种节省空间的解决方案是使用同一数组处理圆形数组。如果仔细观察数组,则在第n个索引之后,下一个索引始终从0开始,因此如果使用(i)%n,则使用mod运算符,我们可以轻松访问循环列表的元素并运行从第i个索引到第n + i个索引的循环,并应用mod,我们可以在给定数组内的圆形数组中进行遍历,而无需使用任何额外的空间。
下面是上述方法的实现:
C++
// C++ program to demonstrate the use of circular
// array without using extra memory space
#include
using namespace std;
// Function to find the Next Greater Element(NGE)
void printNGE(int A[], int n)
{
int k;
for (int i = 0; i < n; i++) {
// Initialise k as -1 which is printed when no NGE
// found
k = -1; //
for (int j = i + 1; j < n + i; j++) {
if (A[i] < A[j % n]) {
printf("%d ", A[j % n]);
k = 1;
break;
}
}
if (k == -1) // Gets executed when no NGE found
printf("-1 ");
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 8, 6, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
printNGE(arr, N);
return 0;
}
Java
// Java program to demonstrate the
// use of circular array without
// using extra memory space
import java.io.*;
class GFG{
// Function to find the Next
// Greater Element(NGE)
static void printNGE(int A[], int n)
{
int k;
for(int i = 0; i < n; i++)
{
// Initialise k as -1 which is
// printed when no NGE found
k = -1;
for(int j = i + 1; j < n + i; j++)
{
if (A[i] < A[j % n])
{
System.out.print(A[j % n] + " ");
k = 1;
break;
}
}
// Gets executed when no NGE found
if (k == -1)
System.out.print("-1 ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int[] arr = { 8, 6, 7 };
int N = arr.length;
// Function call
printNGE(arr, N);
}
}
// This code is contributed by yashbeersingh42
Python3
# Python3 program to demonstrate the use of circular
# array without using extra memory space
# Function to find the Next Greater Element(NGE)
def printNGE(A, n) :
for i in range(n) :
# Initialise k as -1 which is printed when no NGE
# found
k = -1
for j in range(i + 1, n + i) :
if (A[i] < A[j % n]) :
print(A[j % n], end = " ")
k = 1
break
if (k == -1) : # Gets executed when no NGE found
print("-1 ", end = "")
# Given array arr[]
arr = [ 8, 6, 7 ]
N = len(arr)
# Function call
printNGE(arr, N)
# This code is contributed by divyeshrabadia07
C#
// C# program to demonstrate the
// use of circular array without
// using extra memory space
using System;
class GFG {
// Function to find the Next
// Greater Element(NGE)
static void printNGE(int[] A, int n)
{
int k;
for(int i = 0; i < n; i++)
{
// Initialise k as -1 which is
// printed when no NGE found
k = -1;
for(int j = i + 1; j < n + i; j++)
{
if (A[i] < A[j % n])
{
Console.Write(A[j % n] + " ");
k = 1;
break;
}
}
// Gets executed when no NGE found
if (k == -1)
Console.Write("-1 ");
}
}
static void Main()
{
// Given array arr[]
int[] arr = { 8, 6, 7 };
int N = arr.Length;
// Function call
printNGE(arr, N);
}
}
// This code is contributed by divyesh072019
-1 7 8
时间复杂度: O(n 2 )
辅助空间: O(1)
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