程序检查给定的数字是否幸运(所有数字都不相同)

如果数字的所有数字都不相同，那么数字是幸运的。如何检查给定的数字是否幸运。
例子：

Input: n = 983
Output: true
All digits are different

Input: n = 9838
Output: false
8 appears twice

强烈建议您最小化浏览器，然后自己尝试。
想法是遍历给定数字的每个数字，并将遍历的数字标记为已访问。由于数字的总数为10，因此我们需要一个大小仅为10的布尔数组来标记访问的数字。
以下是上述想法的实现。

C++

// C++ program to check if a given number is lucky
#include
using namespace std;
 
// This function returns true if n is lucky
bool isLucky(int n)
{
    // Create an array of size 10 and initialize all
    // elements as false. This array is used to check
    // if a digit is already seen or not.
    bool arr[10];
    for (int i=0; i<10; i++)
        arr[i] = false;
 
    // Traverse through all digits of given number
    while (n > 0)
    {
        // Find the last digit
        int digit = n%10;
 
        // If digit is already seen, return false
        if (arr[digit])
           return false;
 
        // Mark this digit as seen
        arr[digit] = true;
 
        // REmove the last digit from number
        n = n/10;
    }
    return true;
}
 
// Driver program to test above function.
int main()
{
    int arr[] = {1291, 897, 4566, 1232, 80, 700};
    int n = sizeof(arr)/sizeof(arr[0]);
 
    for (int i=0; i

---

Java

// Java program to check if
// a given number is lucky
 
class GFG
{
    // This function returns true if n is lucky
    static boolean isLucky(int n)
    {
        // Create an array of size 10 and initialize all
        // elements as false. This array is used to check
        // if a digit is already seen or not.
        boolean arr[]=new boolean[10];
        for (int i = 0; i < 10; i++)
            arr[i] = false;
     
        // Traverse through all digits
        // of given number
        while (n > 0)
        {
            // Find the last digit
            int digit = n % 10;
     
            // If digit is already seen,
            // return false
            if (arr[digit])
            return false;
     
            // Mark this digit as seen
            arr[digit] = true;
     
            // Remove the last digit from number
            n = n / 10;
        }
        return true;
    }
     
    // Driver code
    public static void main (String[] args)
    {
    int arr[] = {1291, 897, 4566, 1232, 80, 700};
        int n = arr.length;
     
        for (int i = 0; i < n; i++)
            if(isLucky(arr[i]))
                System.out.print(arr[i] + " is Lucky \n");
            else
            System.out.print(arr[i] + " is not Lucky \n");
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# python program to check if a
# given number is lucky
 
import math
 
# This function returns true
# if n is lucky
def isLucky(n):
     
    # Create an array of size 10
    # and initialize all elements
    # as false. This array is
    # used to check if a digit
    # is already seen or not.
    ar = [0] * 10
     
    # Traverse through all digits
    # of given number
    while (n > 0):
         
        #Find the last digit
        digit = math.floor(n % 10)
 
        # If digit is already seen,
        # return false
        if (ar[digit]):
            return 0
 
        # Mark this digit as seen
        ar[digit] = 1
 
        # REmove the last digit
        # from number
        n = n / 10
     
    return 1
 
# Driver program to test above function.
arr = [1291, 897, 4566, 1232, 80, 700]
n = len(arr)
 
for i in range(0, n):
    k = arr[i]
    if(isLucky(k)):
        print(k, " is Lucky ")
    else:
        print(k, " is not Lucky ")
     
# This code is contributed by Sam007.

C#

// C# program to check if
// a given number is lucky
using System;
 
class GFG {
     
    // This function returns true if
    // n is lucky
    static bool isLucky(int n)
    {
         
        // Create an array of size 10
        // and initialize all elements
        // as false. This array is used
        // to check if a digit is
        // already seen or not.
        bool []arr = new bool[10];
         
        for (int i = 0; i < 10; i++)
            arr[i] = false;
     
        // Traverse through all digits
        // of given number
        while (n > 0)
        {
            // Find the last digit
            int digit = n % 10;
     
            // If digit is already seen,
            // return false
            if (arr[digit])
                return false;
     
            // Mark this digit as seen
            arr[digit] = true;
     
            // Remove the last digit
            // from number
            n = n / 10;
        }
         
        return true;
    }
     
    // Driver code
    public static void Main ()
    {
    int []arr = {1291, 897, 4566, 1232,
                               80, 700};
        int n = arr.Length;
     
        for (int i = 0; i < n; i++)
            if(isLucky(arr[i]))
                Console.Write(arr[i] +
                        " is Lucky \n");
            else
            Console.Write(arr[i] +
                    " is not Lucky \n");
    }
}
 
// This code is contributed by sam007.

PHP

 0)
    {
        // Find the last digit
        $digit = $n % 10;
 
        // If digit is already seen,
        // return false
        if ($arr[$digit])
        return false;
 
        // Mark this digit as seen
        $arr[$digit] = true;
 
        // Remove the last digit
        // from number
        $n = (int)($n / 10);
    }
    return true;
}
 
// Driver Code
$arr = array(1291, 897, 4566,
             1232, 80, 700);
$n = sizeof($arr);
 
for ($i = 0; $i < $n; $i++)
    if(isLucky($arr[$i]))
        echo $arr[$i] , " is Lucky \n";
    else
        echo $arr[$i] , " is not Lucky \n";
 
// This code is contributed by jit_t
?>

Javascript

 
// Javascript program to check if a given number is lucky
 
// This function returns true if n is lucky
function isLucky(n)
{
    // Create an array of size 10 and initialize all
    // elements as false. This array is used to check
    // if a digit is already seen or not.
    var arr=Array(10).fill(0);
    for (var i=0; i<10; i++)
        arr[i] = false;
 
    // Traverse through all digits of given number
    while (n > 0)
    {
        // Find the last digit
        var digit = n%10;
 
        // If digit is already seen, return false
        if (arr[digit])
        return false;
 
        // Mark this digit as seen
        arr[digit] = true;
 
        // REmove the last digit from number
        n = parseInt(n/10);
    }
    return true;
}
 
// Driver program to test above function.
var arr = [1291, 897, 4566, 1232, 80, 700]
var n = arr.length;
for (var i=0; i<n; i++)
    isLucky(arr[i])? document.write( arr[i] + " is Lucky<br>"):
                    document.write(arr[i] + " is not Lucky<br>");
 

输出：

1291 is not Lucky
897 is Lucky
4566 is not Lucky
1232 is not Lucky
80 is Lucky
700 is not Lucky