检查一个号码是否以另一个号码结尾

给定两个数字A和B ，其中( A> B )，任务是检查B是否为A的后缀。如果是后缀，则打印“是”；否则，则打印“否” 。

例子：

Input: A = 12345, B = 45
Output: Yes

Input: A = 12345, B = 123
Output: No

方法1：

1. 将给定的数字A和B分别转换为字符串str1和str2 。
2. 从字符串的末尾遍历两个字符串。
3. 在遍历字符串，如果来自str1和str2的任何索引字符都不相等，则打印“ No” 。
4. 否则打印“是”。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
  
// Function to check if B is a
// suffix of A or not
bool checkSuffix(int A, int B)
{
  
    // Convert numbers into strings
    string s1 = to_string(A);
    string s2 = to_string(B);
  
    // Find the lengths of strings
    // s1 and s2
    int n1 = s1.length();
    int n2 = s2.length();
  
    // Base Case
    if (n1 < n2) {
        return false;
    }
  
    // Traverse the strings s1 & s2
    for (int i = 0; i < n2; i++) {
  
        // If at any index characters
        // are unequals then return false
        if (s1[n1 - i - 1]
            != s2[n2 - i - 1]) {
            return false;
        }
    }
  
    // Return true
    return true;
}
  
// Driver Code
int main()
{
    // Given numbers
    int A = 12345, B = 45;
  
    // Function Call
    bool result = checkSuffix(A, B);
  
    // If B is a suffix of A, then
    // print "Yes"
    if (result) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
// Function to check if B   
// is a suffix of A or not 
public static boolean checkSuffix(int A, 
                                  int B) 
{ 
      
    // Convert numbers into strings 
    String s1 = String.valueOf(A); 
    String s2 = String.valueOf(B); 
      
    // Find the lengths of strings 
    // s1 and s2 
    int n1 = s1.length(); 
    int n2 = s2.length(); 
      
    // Base case 
    if (n1 < n2)
    { 
        return false; 
    } 
      
    // Traverse the strings s1 & s2 
    for(int i = 0; i < n2; i++) 
    { 
          
       // If at any index characters 
       // are unequals then return false 
       if (s1.charAt(n1 - i - 1) != 
           s2.charAt(n2 - i - 1))
       { 
           return false; 
       } 
    } 
      
    // Return true 
    return true; 
} 
  
// Driver code
public static void main(String[] args)
{
          
    // Given numbers 
    int A = 12345, B = 45; 
      
    // Function Call 
    boolean result = checkSuffix(A, B); 
      
    // If B is a suffix of A,  
    // then print "Yes" 
    if (result)
    { 
        System.out.print("Yes");
    } 
    else
    { 
        System.out.println("No");
    } 
}
}
  
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach 
  
# Function to check if B is a 
# suffix of A or not 
def checkSuffix(A, B):
  
    # Convert numbers into strings 
    s1 = str(A); 
    s2 = str(B); 
  
    # Find the lengths of strings 
    # s1 and s2 
    n1 = len(s1) 
    n2 = len(s2) 
  
    # Base Case 
    if (n1 < n2): 
        return False; 
      
    # Traverse the strings s1 & s2 
    for i in range(n2): 
  
        # If at any index characters 
        # are unequals then return false 
        if (s1[n1 - i - 1] != s2[n2 - i - 1]):
            return False; 
              
    # Return true 
    return True; 
      
# Driver Code 
  
# Given numbers 
A = 12345
B = 45; 
  
# Function Call 
result = checkSuffix(A, B); 
  
# If B is a suffix of A, then 
# print "Yes" 
if (result): 
    print("Yes")
else: 
    print("No")
  
# This code is contributed by grand_master

C#

// C# program for the above approach 
using System;
class GFG{ 
      
// Function to check if B 
// is a suffix of A or not 
public static bool checkSuffix(int A, 
                               int B) 
{ 
      
    // Convert numbers into strings 
    string s1 = A.ToString(); 
    string s2 = B.ToString(); 
      
    // Find the lengths of strings 
    // s1 and s2 
    int n1 = s1.Length; 
    int n2 = s2.Length; 
      
    // Base case 
    if (n1 < n2) 
    { 
        return false; 
    } 
      
    // Traverse the strings s1 & s2 
    for(int i = 0; i < n2; i++) 
    { 
          
        // If at any index characters 
        // are unequals then return false 
        if (s1[n1 - i - 1] !=  s2[n2 - i - 1])
        { 
            return false; 
        } 
    } 
      
    // Return true 
    return true; 
} 
  
// Driver code 
public static void Main(string[] args) 
{ 
          
    // Given numbers 
    int A = 12345, B = 45; 
      
    // Function Call 
    bool result = checkSuffix(A, B); 
      
    // If B is a suffix of A, 
    // then print "Yes" 
    if (result) 
    { 
        Console.Write("Yes"); 
    } 
    else
    { 
        Console.Write("No"); 
    } 
} 
} 
  
// This code is contributed by rutvik_56

输出：

Yes

方法2：使用内置函数std :: boost :: algorithm :: ends_with() (包含在C++ Boost库中)，该函数用于检查任何字符串是否包含另一个字符串的后缀。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
#include 
using namespace std;
  
// Function to check if B is a
// suffix of A or not
void checkSuffix(int A, int B)
{
  
    // Convert numbers into strings
    string s1 = to_string(A);
    string s2 = to_string(B);
  
    bool result;
  
    // Check if s2 is a suffix of s1
    // or not using ends_with() function
    result = boost::algorithm::ends_with(s1,
                                         s2);
  
    // If result is true, print "Yes"
    if (result) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}
  
// Driver Code
int main()
{
    // Given numbers
    int A = 12345, B = 45;
  
    // Function Call
    checkSuffix(A, B);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to check if B is a
// suffix of A or not
static void checkSuffix(int A, int B)
{
  
    // Convert numbers into Strings
    String s1 = String.valueOf(A);
    String s2 = String.valueOf(B);
  
    boolean result;
  
    // Check if s2 is a suffix of s1
    // or not 
    result = s1.endsWith(s2);
  
    // If result is true, print "Yes"
    if (result)
    {
        System.out.print("Yes");
    }
    else 
    {
        System.out.print("No");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    // Given numbers
    int A = 12345, B = 45;
  
    // Function Call
    checkSuffix(A, B);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
  
# Function to check if B is 
# a suffix of A or not
def checkSuffix(A, B):
  
    # Convert numbers into strings
    s1 = str(A)
    s2 = str(B)
  
    # Check if s2 is a suffix of s1
    # or not 
    result = s1.endswith(s2)
  
    # If result is true print "Yes"
    if (result):
        print("Yes")
    else:
        print("No")
  
# Driver code
if __name__ == '__main__':
  
    # Given numbers
    A = 12345
    B = 45
  
    # Function call
    checkSuffix(A, B)
  
# This code is contributed by himanshu77

C＃

// C# program for the above approach
using System;
  
class GFG{
  
// Function to check if B is a
// suffix of A or not
static void checkSuffix(int A, int B)
{
  
    // Convert numbers into Strings
    String s1 = String.Join("", A);
    String s2 = String.Join("", B);
  
    bool result;
  
    // Check if s2 is a suffix of s1
    // or not 
    result = s1.EndsWith(s2);
  
    // If result is true, print "Yes"
    if (result)
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    // Given numbers
    int A = 12345, B = 45;
  
    // Function Call
    checkSuffix(A, B);
}
}
  
// This code is contributed by 29AjayKumar

输出：

Yes

方法3：

1. 从A减去B。
2. 通过使用本文讨论的方法3 ，找到B(例如X )中的位数。
3. 检查A是否以至少X个数字零结尾。如果是，则打印“是” 。
4. 否则打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
#include 
using namespace std;
  
// Function to check if B is a
// suffix of A or not
bool checkSuffix(int A, int B)
{
  
    // Find the number of digit in B
    int digit_B = log10(B) + 1;
  
    // Subtract B from A
    A -= B;
  
    // Returns true,
    // if B is a suffix of A
    return (A % int(pow(10, digit_B)));
}
  
// Driver Code
int main()
{
    // Given numbers
    int A = 12345, B = 45;
  
    // Function Call
    bool result = checkSuffix(A, B);
  
    // If B is a suffix of A, then
    // print "Yes"
    if (!result) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to check if B 
// is a suffix of A or not
static boolean checkSuffix(int A, int B)
{
  
    // Find the number of digit in B
    int digit_B = (int) (Math.log10(B) + 1);
  
    // Subtract B from A
    A -= B;
  
    // Returns true,
    // if B is a suffix of A
    return (A % (int)(Math.pow(10, digit_B)) > 0);
}
  
// Driver code
public static void main(String[] args)
{
  
    // Given numbers
    int A = 12345, B = 45;
  
    // Function call
    boolean result = checkSuffix(A, B);
  
    // If B is a suffix of A, 
    // then print "Yes"
    if (!result)
    {
        System.out.print("Yes");
    }
    else 
    {
        System.out.print("No");
    }
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
import math
  
# Function to check if B is a
# suffix of A or not
def checkSuffix(A, B):
  
    # Find the number of digit in B
    digit_B = int(math.log10(B)) + 1;
  
    # Subtract B from A
    A -= B;
  
    # Returns true,
    # if B is a suffix of A
    return (A % int(math.pow(10, digit_B)));
  
# Driver Code
  
# Given numbers
A = 12345; B = 45;
  
# Function Call
result = checkSuffix(A, B);
  
# If B is a suffix of A, then
# print "Yes"
if (result == 0):
    print("Yes");
  
else: 
    print("No");
  
# This code is contributed by Nidhi_biet

C＃

// C# program for the above approach
using System;
class GFG{
  
// Function to check if B 
// is a suffix of A or not
static bool checkSuffix(int A, int B)
{
  
    // Find the number of digit in B
    int digit_B = (int)(Math.Log10(B) + 1);
  
    // Subtract B from A
    A -= B;
  
    // Returns true,
    // if B is a suffix of A
    return (A % (int)(Math.Pow(10, digit_B)) > 0);
}
  
// Driver code
public static void Main()
{
  
    // Given numbers
    int A = 12345, B = 45;
  
    // Function call
    bool result = checkSuffix(A, B);
  
    // If B is a suffix of A, 
    // then print "Yes"
    if (!result)
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
  
// This code is contributed by Code_Mech

输出：

Yes

时间复杂度： O(1)