删除数字的第N位的程序

给定数字num和数字n，任务是从开头和结尾删除数字num的第n个数字。
例子：

Input: num = 1234, n = 3 
Output: num_after_deleting_from_starting = 124, num_after_deleting_from_end = 134
Input: num = 4516312, n = 2 
Output: num_after_deleting_from_starting = 416312, num_after_deleting_from_end = 451632

方法：

• 要从开始删除第n个数字，请执行以下操作：
  1. 获取要删除的数字和第n个数字。
  2. 计算位数
  3. 通过使用变量i进行计数，循环位数的时间。
  4. 如果i等于(位数– i)，则跳过，否则将第i位添加为[new_number =(new_number * 10)+ ith_digit]。
• 要从结尾删除第n个数字：
  1. 获取要删除的数字和第n个数字。
  2. 通过使用变量i进行计数，循环位数的时间。
  3. 如果i等于(n)，则跳过，否则将第i个数字添加为[new_number =(new_number * 10)+ ith_digit]。

执行：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to delete nth digit
// from starting
int deleteFromStart(int num, int n)
{
 
    // Get the number of digits
    int d = log10(num) + 1;
 
    // Declare a variable
    // to form the reverse resultant number
    int rev_new_num = 0;
 
    // Loop with the number
    for (int i = 0; num != 0; i++) {
 
        int digit = num % 10;
        num = num / 10;
 
        if (i == (d - n)) {
            continue;
        }
        else {
 
            rev_new_num = (rev_new_num * 10) + digit;
        }
    }
 
    // Declare a variable
    // to form the resultant number
    int new_num = 0;
 
    // Loop with the number
    for (int i = 0; rev_new_num != 0; i++) {
 
        new_num = (new_num * 10)
                  + (rev_new_num % 10);
        rev_new_num = rev_new_num / 10;
    }
 
    // Return the resultant number
    return new_num;
}
 
// Function to delete nth digit
// from ending
int deleteFromEnd(int num, int n)
{
 
    // Declare a variable
    // to form the reverse resultant number
    int rev_new_num = 0;
 
    // Loop with the number
    for (int i = 1; num != 0; i++) {
 
        int digit = num % 10;
        num = num / 10;
 
        if (i == n) {
            continue;
        }
        else {
 
            rev_new_num = (rev_new_num * 10) + digit;
        }
    }
 
    // Declare a variable
    // to form the resultant number
    int new_num = 0;
 
    // Loop with the number
    for (int i = 0; rev_new_num != 0; i++) {
 
        new_num = (new_num * 10)
                  + (rev_new_num % 10);
        rev_new_num = rev_new_num / 10;
    }
 
    // Return the resultant number
    return new_num;
}
 
// Driver code
int main()
{
 
    // Get the number
    int num = 1234;
    cout << "Number: " << num << endl;
 
    // Get the digit number to be deleted
    int n = 3;
    cout << "Digit to be deleted: " << n << endl;
 
    // Remove the nth digit from starting
    cout << "Number after " << n
         << " digit deleted from starting: "
         << deleteFromStart(num, n) << endl;
 
    // Remove the nth digit from ending
    cout << "Number after " << n
         << " digit deleted from ending: "
         << deleteFromEnd(num, n) << endl;
 
    return 0;
}

Java

// Java implementation of above approach
 
 
class GFG
{
    // Function to delete nth digit
    // from starting
    static int deleteFromStart(int num, int n)
    {
     
        // Get the number of digits
        int d = (int)Math.log10(num) + 1;
     
        // Declare a variable
        // to form the reverse resultant number
        int rev_new_num = 0;
     
        // Loop with the number
        for (int i = 0; num != 0; i++) {
     
            int digit = num % 10;
            num = num / 10;
     
            if (i == (d - n)) {
                continue;
            }
            else {
     
                rev_new_num = (rev_new_num * 10) + digit;
            }
        }
     
        // Declare a variable
        // to form the resultant number
        int new_num = 0;
     
        // Loop with the number
        for (int i = 0; rev_new_num != 0; i++) {
     
            new_num = (new_num * 10)
                    + (rev_new_num % 10);
            rev_new_num = rev_new_num / 10;
        }
     
        // Return the resultant number
        return new_num;
    }
     
    // Function to delete nth digit
    // from ending
    static int deleteFromEnd(int num, int n)
    {
     
        // Declare a variable
        // to form the reverse resultant number
        int rev_new_num = 0;
     
        // Loop with the number
        for (int i = 1; num != 0; i++) {
     
            int digit = num % 10;
            num = num / 10;
     
            if (i == n) {
                continue;
            }
            else {
     
                rev_new_num = (rev_new_num * 10) + digit;
            }
        }
     
        // Declare a variable
        // to form the resultant number
        int new_num = 0;
     
        // Loop with the number
        for (int i = 0; rev_new_num != 0; i++) {
     
            new_num = (new_num * 10)
                    + (rev_new_num % 10);
            rev_new_num = rev_new_num / 10;
        }
     
        // Return the resultant number
        return new_num;
    }
     
    // Driver code
    public static void main(String []args)
    {
     
        // Get the number
        int num = 1234;
        System.out.println("Number: " + num );
     
        // Get the digit number to be deleted
        int n = 3;
        System.out.println("Digit to be deleted: " + n );
     
        // Remove the nth digit from starting
        System.out.println("Number after " + n
            + " digit deleted from starting: "
            + deleteFromStart(num, n));
     
        // Remove the nth digit from ending
        System.out.println( "Number after " + n
            + " digit deleted from ending: "
            + deleteFromEnd(num, n));
     
         
    }
 
}
 
// This code is contributed by ihritik

Python3

# Python3 implementation of above approach
 
# Function to delete nth digit
# from starting
import math;
def deleteFromStart(num, n):
 
    # Get the number of digits
    d = (math.log10(num) + 1);
 
    # Declare a variable to form
    # the reverse resultant number
    rev_new_num = 0;
 
    # Loop with the number
    i = 0;
    while (num != 0):
 
        digit = num % 10;
        num = int(num / 10);
 
        if (i != (int(d) - n)):
            rev_new_num = ((rev_new_num * 10) +
                                        digit);
        i += 1;
 
    # Declare a variable to form the
    # resultant number
    new_num = 0;
 
    # Loop with the number
    i = 0;
    while (rev_new_num != 0):
 
        new_num = ((new_num * 10) +
                   (rev_new_num % 10));
        rev_new_num = int(rev_new_num / 10);
        i += 1;
 
    # Return the resultant number
    return new_num;
 
# Function to delete nth digit
# from ending
def deleteFromEnd(num, n):
 
    # Declare a variable to form
    # the reverse resultant number
    rev_new_num = 0;
 
    # Loop with the number
    i = 1;
    while (num != 0):
 
        digit = num % 10;
        num = int(num / 10);
 
        if (i != n):
            rev_new_num = ((rev_new_num * 10) +
                                        digit);
        i += 1;
 
    # Declare a variable
    # to form the resultant number
    new_num = 0;
 
    # Loop with the number
    i = 0;
    while (rev_new_num != 0):
 
        new_num = ((new_num * 10) +
                   (rev_new_num % 10));
        rev_new_num = int(rev_new_num / 10);
        i += 1;
 
    # Return the resultant number
    return new_num;
 
# Driver code
# Get the number
num = 1234;
print("Number:", num);
 
# Get the digit number to be deleted
n = 3;
print("Digit to be deleted:", n);
 
# Remove the nth digit from starting
print("Number after", n,
      "digit deleted from starting:",
            deleteFromStart(num, n));
 
# Remove the nth digit from ending
print("Number after", n,
      "digit deleted from ending:",
            deleteFromEnd(num, n));
 
# This code is contributed by chandan_jnu

C#

// C# implementation of the above approach
 
using System;
class GFG
{
    // Function to delete nth digit
    // from starting
    static int deleteFromStart(int num, int n)
    {
     
        // Get the number of digits
        int d = (int)Math.Log10(num) + 1;
     
        // Declare a variable
        // to form the reverse resultant number
        int rev_new_num = 0;
     
        // Loop with the number
        for (int i = 0; num != 0; i++) {
     
            int digit = num % 10;
            num = num / 10;
     
            if (i == (d - n)) {
                continue;
            }
            else {
     
                rev_new_num = (rev_new_num * 10) + digit;
            }
        }
     
        // Declare a variable
        // to form the resultant number
        int new_num = 0;
     
        // Loop with the number
        for (int i = 0; rev_new_num != 0; i++) {
     
            new_num = (new_num * 10)
                    + (rev_new_num % 10);
            rev_new_num = rev_new_num / 10;
        }
     
        // Return the resultant number
        return new_num;
    }
     
    // Function to delete nth digit
    // from ending
    static int deleteFromEnd(int num, int n)
    {
     
        // Declare a variable
        // to form the reverse resultant number
        int rev_new_num = 0;
     
        // Loop with the number
        for (int i = 1; num != 0; i++) {
     
            int digit = num % 10;
            num = num / 10;
     
            if (i == n) {
                continue;
            }
            else {
     
                rev_new_num = (rev_new_num * 10) + digit;
            }
        }
     
        // Declare a variable
        // to form the resultant number
        int new_num = 0;
     
        // Loop with the number
        for (int i = 0; rev_new_num != 0; i++) {
     
            new_num = (new_num * 10)
                    + (rev_new_num % 10);
            rev_new_num = rev_new_num / 10;
        }
     
        // Return the resultant number
        return new_num;
    }
     
    // Driver code
    public static void Main()
    {
     
        // Get the number
        int num = 1234;
        Console.WriteLine("Number: " + num );
     
        // Get the digit number to be deleted
        int n = 3;
        Console.WriteLine("Digit to be deleted: " + n );
     
        // Remove the nth digit from starting
        Console.WriteLine("Number after " + n
            + " digit deleted from starting: "
            + deleteFromStart(num, n));
     
        // Remove the nth digit from ending
        Console.WriteLine( "Number after " + n
            + " digit deleted from ending: "
            + deleteFromEnd(num, n));
     
         
    }
 
}
 
// This code is contributed by ihritik

PHP

// This code is contributed by jit_t.

Javascript

 
    // Javascript implementation of
    // the above approach
     
    // Function to delete nth digit
    // from starting
    function deleteFromStart(num, n)
    {
       
        // Get the number of digits
        let d = parseInt(Math.log10(num), 10) + 1;
       
        // Declare a variable
        // to form the reverse resultant number
        let rev_new_num = 0;
       
        // Loop with the number
        for (let i = 0; num != 0; i++) {
       
            let digit = num % 10;
            num = parseInt(num / 10, 10);
       
            if (i == (d - n)) {
                continue;
            }
            else {
       
                rev_new_num = (rev_new_num * 10) + digit;
            }
        }
       
        // Declare a variable
        // to form the resultant number
        let new_num = 0;
       
        // Loop with the number
        for (let i = 0; rev_new_num != 0; i++) {
       
            new_num = (new_num * 10)
                    + (rev_new_num % 10);
            rev_new_num = parseInt(rev_new_num / 10, 10);
        }
       
        // Return the resultant number
        return new_num;
    }
       
    // Function to delete nth digit
    // from ending
    function deleteFromEnd(num, n)
    {
       
        // Declare a variable
        // to form the reverse resultant number
        let rev_new_num = 0;
       
        // Loop with the number
        for (let i = 1; num != 0; i++) {
       
            let digit = num % 10;
            num = parseInt(num / 10, 10);
       
            if (i == n) {
                continue;
            }
            else {
       
                rev_new_num = (rev_new_num * 10) + digit;
            }
        }
       
        // Declare a variable
        // to form the resultant number
        let new_num = 0;
       
        // Loop with the number
        for (let i = 0; rev_new_num != 0; i++) {
       
            new_num = (new_num * 10)
                    + (rev_new_num % 10);
            rev_new_num = parseInt(rev_new_num / 10, 10);
        }
       
        // Return the resultant number
        return new_num;
    }
     
    // Get the number
    let num = 1234;
    document.write("Number: " + num + "</br>");
 
    // Get the digit number to be deleted
    let n = 3;
    document.write("Digit to be deleted: " + n  + "</br>");
 
    // Remove the nth digit from starting
    document.write("Number after " + n
                      + " digit deleted from starting: "
                      + deleteFromStart(num, n) + "</br>");
 
    // Remove the nth digit from ending
    document.write( "Number after " + n
                      + " digit deleted from ending: "
                      + deleteFromEnd(num, n));
     

输出：

Number: 1234
Digit to be deleted: 3
Number after 3 digit deleted from starting: 124
Number after 3 digit deleted from ending: 134

另一种方法：(将数字转换为字符串)
时间复杂度： O(1)

C++

// C++ implementation to delete nth digit
// from starting with O(1) time complexity.
#include
using namespace std;
 
// function to delete nth number from starting
static string fromStart(string inp, int del)
{
    string inp1 = inp.substr(0, del - 1);
    string inp2 = inp.substr(del, inp.length());
    return inp1 + inp2;
}
 
// function to delete nth number from ending
static string fromEnd(string inp, int del)
{
    string inp1 = inp.substr(0, inp.length() - del);
    string inp2 = inp.substr(inp.length() - del + 1,
                                      inp.length());
    return inp1 + inp2;
}
 
// Driver Code
int main()
{
    int in = 1234;
     
    // type cast input number to string
    stringstream ss;
    ss << in;
    string inp = ss.str();
    int del = 3;
    cout << "num_after_deleting_from_starting "
         << fromStart(inp, del) << endl;
    cout << "num_after_deleting_from_ending "
         << fromEnd(inp, del) << endl;
    return 0;
}
 
// This code is contributed by chandan_jnu

Java

// Java implementation to delete nth digit
// from starting with O(1) time complexity.
 
public class DeleteN {
 
    public static void main(String args[]) {
 
        int in = 1234;
        // type cast input number to string
        String inp = Integer.toString(in);
        int del = 3;
        System.out.println("num_after_deleting_from_starting " + fromStart(inp, del));
        System.out.println("num_after_deleting_from_ending " + fromEnd(inp, del));
    }
 
    // function to delete nth number from starting
    static String fromStart(String inp, int del) {
 
        try {
            String inp1 = inp.substring(0, del - 1);
            String inp2 = inp.substring(del, inp.length());
            return inp1 + inp2;
        }
 
        catch (Exception e) {
            return "Check Input";
        }
    }
 
    // function to delete nth number from ending
    static String fromEnd(String inp, int del) {
 
        try {
            String inp1 = inp.substring(0, inp.length() - del);
            String inp2 = inp.substring(inp.length() - del + 1, inp.length());
            return inp1 + inp2;
        }
 
        catch (Exception e) {
            return "Check Input";
        }
    }
 
}

Python3

# Python3 implementation to delete nth digit
# from starting with O(1) time complexity.
 
# function to del1ete nth number
# from starting
def fromStart(inp, del11):
 
    inp1 = inp[0:del1 - 1];
    inp2 = inp[del1:len(inp)];
    return inp1 + inp2;
 
# function to delete nth number
# from ending
def fromEnd(inp, del1):
    inp1 = inp[0:len(inp) - del1];
    inp2 = inp[len(inp) - del1 + 1:len(inp)];
    return inp1 + inp2;
 
# Driver Code
in1 = 1234;
 
# type cast input number to string
inp = str(in1);
del1 = 3;
print("num_after_deleting_from_starting",
                   fromStart(inp, del1));
print("num_after_deleting_from_ending",
                   fromEnd(inp, del1));
 
# This code is contributed by chandan_jnu

C＃

// C# implementation to delete nth digit
// from starting with O(1) time complexity.
using System ;
 
public class DeleteN {
 
    public static void Main() {
 
        int num = 1234;
         
        // type cast input number to string
        string inp = Convert.ToString(num) ;
        int del = 3;
        Console.WriteLine("num_after_deleting_from_starting "
                            + fromStart(inp, del));
        Console.WriteLine("num_after_deleting_from_ending "
                            + fromEnd(inp, del));
    }
 
    // function to delete nth number from starting
    static String fromStart(string inp, int del) {
 
        try {
            string inp1 = inp.Substring(0, del - 1);
            string inp2 = inp.Substring(del, inp.Length - del);
            return inp1 + inp2;
        }
 
        catch (Exception ) {
            return "Check Input";
        }
    }
 
    // function to delete nth number from ending
    static String fromEnd(string inp, int del) {
 
        try {
            string inp1 = inp.Substring(0, inp.Length - del);
            string inp2 = inp.Substring(inp.Length - del + 1, del - 1);
            return inp1 + inp2;
        }
 
        catch (Exception e) {
            Console.WriteLine(e) ;
            return "Check Input";
        }
    }
}
 
// This code is contributed by Ryuga

输出：

num_after_deleting_from_starting 124
num_after_deleting_from_ending 134