检查数组是否是从1到N的数字排列

给定一个包含N个正整数的数组arr ，任务是检查给定的数组arr是否表示一个排列。

A sequence of N integers is called a permutation if it contains all integers from 1 to N exactly once.

例子：

Input: arr[] = {1, 2, 5, 3, 2}
Output: No
Explanation:
The given array is not a permutation of numbers from 1 to N, because it contains 2 twice, and 4 is missing for the array to represent a permutation of length 5.

Input: arr[] = {1, 2, 5, 3, 4}
Output: Yes
Explanation:
Given array contains all integers from 1 to 5 exactly once. Hence, it represents a permutation of length 5.

天真的方法：显然，给定的数组将仅表示长度N的排列，其中N是数组的长度。因此，我们必须在给定数组中搜索从1到N的每个元素。如果找到了所有元素，则数组代表一个排列，否则就没有排列。

时间复杂度： O(N ² )

高效方法：
可以使用设置的数据结构来优化上述方法。

1. 遍历给定的数组，并将每个元素插入设置的数据结构中。
2. 另外，在数组中找到最大元素。该最大元素将是值N，它将代表集合的大小。
3. 遍历数组后，检查集合的大小是否等于N。
4. 如果集合的大小等于N，则数组表示一个排列，否则不排列。

下面是上述方法的实现：

C++

// C++ Program to decide if an
// array represents a permutation or not
  
#include 
using namespace std;
  
// Function to check if an
// array represents a permutation or not
bool permutation(int arr[], int n)
{
    // Set to check the count
    // of non-repeating elements
    set hash;
  
    int maxEle = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Insert all elements in the set
        hash.insert(arr[i]);
  
        // Calculating the max element
        maxEle = max(maxEle, arr[i]);
    }
  
    if (maxEle != n)
        return false;
  
    // Check if set size is equal to n
    if (hash.size() == n)
        return true;
  
    return false;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 5, 3, 2 };
    int n = sizeof(arr) / sizeof(int);
  
    if (permutation(arr, n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
  
    return 0;
}

Java

// Java Program to decide if an
// array represents a permutation or not
import java.util.*;
  
class GFG{
  
// Function to check if an
// array represents a permutation or not
static boolean permutation(int []arr, int n)
{
    // Set to check the count
    // of non-repeating elements
    Set hash = new HashSet(); 
  
    int maxEle = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Insert all elements in the set
        hash.add(arr[i]);
  
        // Calculating the max element
        maxEle = Math.max(maxEle, arr[i]);
    }
  
    if (maxEle != n)
        return false;
  
    // Check if set size is equal to n
    if (hash.size() == n)
        return true;
  
    return false;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 5, 3, 2 };
    int n = arr.length;
  
    if (permutation(arr, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by Surendra_Gangwar

Python3

# Python3 Program to decide if an
# array represents a permutation or not
  
# Function to check if an
# array represents a permutation or not
def permutation(arr, n):
      
        # Set to check the count
    # of non-repeating elements
    s = set()
  
    maxEle = 0;
  
    for i in range(n):
    
        # Insert all elements in the set
        s.add(arr[i]);
  
        # Calculating the max element
        maxEle = max(maxEle, arr[i]);
      
    if (maxEle != n):
        return False
  
    # Check if set size is equal to n
    if (len(s) == n):
        return True;
  
    return False;
  
# Driver code
if __name__=='__main__': 
  
    arr = [ 1, 2, 5, 3, 2 ]
    n = len(arr)
  
    if (permutation(arr, n)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by Princi Singh

C#

// C# Program to decide if an
// array represents a permutation or not
using System;
using System.Collections.Generic;
  
class GFG{
   
// Function to check if an
// array represents a permutation or not
static bool permutation(int []arr, int n)
{
    // Set to check the count
    // of non-repeating elements
    HashSet hash = new HashSet(); 
   
    int maxEle = 0;
   
    for (int i = 0; i < n; i++) {
   
        // Insert all elements in the set
        hash.Add(arr[i]);
   
        // Calculating the max element
        maxEle = Math.Max(maxEle, arr[i]);
    }
   
    if (maxEle != n)
        return false;
   
    // Check if set size is equal to n
    if (hash.Count == n)
        return true;
   
    return false;
}
   
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 5, 3, 2 };
    int n = arr.Length;
   
    if (permutation(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by Princi Singh

输出：

No