检查给定的字符串包含所有数字

给定一个包含字母数字字符的字符串str ，任务是检查该字符串是否包含从1到9的所有数字。

例子：

Input: str = “Geeks12345for69708”
Output: Yes
All the digits from 0 to 9 are
present in the given string.

Input: str = “Amazing1234”
Output: No

方法：创建一个频率数组以标记从0到9的每个数字的频率。最后，遍历该频率数组，如果给定字符串不存在任何数字，则答案为“否”，否则答案将是“是”。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
const int MAX = 10;
  
// Function that returns true
// if ch is a digit
bool isDigit(char ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}
  
// Function that returns true
// if str contains all the
// digits from 0 to 9
bool allDigits(string str, int len)
{
  
    // To mark the present digits
    bool present[MAX] = { false };
  
    // For every character of the string
    for (int i = 0; i < len; i++) {
  
        // If the current character is a digit
        if (isDigit(str[i])) {
  
            // Mark the current digit as present
            int digit = str[i] - '0';
            present[digit] = true;
        }
    }
  
    // For every digit from 0 to 9
    for (int i = 0; i < MAX; i++) {
  
        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }
  
    return true;
}
  
// Driver code
int main()
{
    string str = "Geeks12345for69708";
    int len = str.length();
  
    if (allDigits(str, len))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
static int MAX = 10;
  
// Function that returns true
// if ch is a digit
static boolean isDigit(char ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}
  
// Function that returns true
// if str contains all the
// digits from 0 to 9
static boolean allDigits(String str, int len)
{
  
    // To mark the present digits
    boolean []present = new boolean[MAX];
  
    // For every character of the String
    for (int i = 0; i < len; i++) 
    {
  
        // If the current character is a digit
        if (isDigit(str.charAt(i)))
        {
  
            // Mark the current digit as present
            int digit = str.charAt(i) - '0';
            present[digit] = true;
        }
    }
  
    // For every digit from 0 to 9
    for (int i = 0; i < MAX; i++)
    {
  
        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }
  
    return true;
}
  
// Driver code
public static void main(String[] args)
{
    String str = "Geeks12345for69708";
    int len = str.length();
  
    if (allDigits(str, len))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
MAX = 10
  
# Function that returns true
# if ch is a digit
def isDigit(ch):
    ch = ord(ch)
    if (ch >= ord('0') and ch <= ord('9')):
        return True
    return False
  
# Function that returns true
# if st contains all the
# digits from 0 to 9
def allDigits(st, le):
  
    # To mark the present digits
    present = [False for i in range(MAX)]
  
    # For every character of the sting
    for i in range(le):
  
        # If the current character is a digit
        if (isDigit(st[i])):
  
            # Mark the current digit as present
            digit = ord(st[i]) - ord('0')
            present[digit] = True
  
    # For every digit from 0 to 9
    for i in range(MAX):
  
        # If the current digit is
        # not present in st
        if (present[i] == False):
            return False
  
    return True
  
# Driver code
st = "Geeks12345for69708"
le = len(st)
  
if (allDigits(st, le)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
static int MAX = 10;
  
// Function that returns true
// if ch is a digit
static bool isDigit(char ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}
  
// Function that returns true
// if str contains all the
// digits from 0 to 9
static bool allDigits(String str, int len)
{
  
    // To mark the present digits
    bool []present = new bool[MAX];
  
    // For every character of the String
    for (int i = 0; i < len; i++) 
    {
  
        // If the current character is a digit
        if (isDigit(str[i]))
        {
  
            // Mark the current digit as present
            int digit = str[i] - '0';
            present[digit] = true;
        }
    }
  
    // For every digit from 0 to 9
    for (int i = 0; i < MAX; i++)
    {
  
        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }
  
    return true;
}
  
// Driver code
public static void Main(String[] args)
{
    String str = "Geeks12345for69708";
    int len = str.Length;
  
    if (allDigits(str, len))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by 29AjayKumar

输出：

Yes