检查数组是否是从 1 到 N 的数字排列：设置 2

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 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: 
The given array contains all integers from 1 to 5 exactly once. Hence, it represents a permutation of length 5. 
 

天真的方法：在 O(N ² ) 时间内
这里提到了这种方法
另一种方法：在 O(N) 时间和 O(N) 空间中
这里提到了这种方法。
高效方法：使用HashTable

1. 创建一个 N 大小的 HashTable 来存储从 1 到 N 的每个数字的频率计数
2. 遍历给定的数组并将每个数字的频率存储在 HashTable 中。
3. 然后遍历 HashTable 并检查从 1 到 N 的所有数字的频率是否为 1。
4. 如果上述条件为真，则打印“是”，否则打印“否”。

下面是上述方法的实现：

CPP

// 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
string permutation(int arr[], int N)
{
 
    int hash[N + 1] = { 0 };
 
    // Counting the frequency
    for (int i = 0; i < N; i++) {
        hash[arr[i]]++;
    }
 
    // Check if each frequency is 1 only
    for (int i = 1; i <= N; i++) {
        if (hash[i] != 1)
            return "No";
    }
 
    return "Yes";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 1, 5, 5, 3 };
    int n = sizeof(arr) / sizeof(int);
    cout << permutation(arr, n) << endl;
 
    return 0;
}

Java

// Java program to decide if an array
// represents a permutation or not
class GFG{
  
// Function to check if an
// array represents a permutation or not
static String permutation(int arr[], int N)
{
  
    int []hash = new int[N + 1];
  
    // Counting the frequency
    for (int i = 0; i < N; i++) {
        hash[arr[i]]++;
    }
  
    // Check if each frequency is 1 only
    for (int i = 1; i <= N; i++) {
        if (hash[i] != 1)
            return "No";
    }
  
    return "Yes";
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 1, 5, 5, 3 };
    int n = arr.length;
    System.out.print(permutation(arr, n) +"\n");
}
}
 
// This code is contributed by Princi Singh

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) :
 
    hash = [0]*(N + 1);
 
    # Counting the frequency
    for i in range(N) :
        hash[arr[i]] += 1;
 
    # Check if each frequency is 1 only
    for i in range(1, N + 1) :
        if (hash[i] != 1) :
            return "No";
 
    return "Yes";
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 1, 5, 5, 3 ];
    n = len(arr);
    print(permutation(arr, n));
 
    # This code is contributed by Yash_R

C#

// C# program to decide if an array
// represents a permutation or not
using System;
 
class GFG{
  
    // Function to check if an
    // array represents a permutation or not
    static string permutation(int []arr, int N)
    {
      
        int []hash = new int[N + 1];
      
        // Counting the frequency
        for (int i = 0; i < N; i++) {
            hash[arr[i]]++;
        }
      
        // Check if each frequency is 1 only
        for (int i = 1; i <= N; i++) {
            if (hash[i] != 1)
                return "No";
        }
      
        return "Yes";
    }
      
    // Driver code
    public static void Main(string[] args)
    {
        int []arr = { 1, 1, 5, 5, 3 };
        int n = arr.Length;
        Console.Write(permutation(arr, n) +"\n");
    }
}
 
// This code is contributed by Yash_R

Javascript

 
// JavaScript program to decide if an array
// represents a permutation or not
 
// Function to check if an
// array represents a permutation or not
function permutation(arr, N)
{
 
    var hash = Array(N+1).fill(0);
 
    // Counting the frequency
    for (var i = 0; i < N; i++) {
        hash[arr[i]]++;
    }
 
    // Check if each frequency is 1 only
    for (var i = 1; i <= N; i++) {
        if (hash[i] != 1)
            return "No";
    }
 
    return "Yes";
}
 
// Driver code
var arr = [1, 1, 5, 5, 3];
var n = arr.length;
document.write( permutation(arr, n));
 

输出：

No

时间复杂度： O(N)
辅助空间复杂度： O(N)

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