给定整数N ，任务是找到以数字N开头和结尾的最小和最大的N位数。
例子：
Input: N = 3
Output:
Smallest Number = 303
Largest Number = 393
Explanation:
303 is the smallest 3 digit number starting and ending with 3.
393 is the largest 3 digit number starting and ending with 3.
Input: N = 1
Output:
Smallest Number = 1
Largest Number = 1
Explanation:
1 is both the smallest and the largest 1 digit number which starts and ends with 1.
方法：
我们知道，最大和最小的N位数字是9999…9 ，其中9重复N次并重复1000…。 0 ，其中0分别重复N-1次。
我们得到的最小和最大的n位数字开始和结束与N，我们需要更换第一和最小的最后一个数字，并用N最大的n个位数。
我们必须注意拐角处的情况，即当N = 1时，最大和最小数目均为1 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find n digit
// largest number starting
// and ending with n
string findNumberL(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
string result = "";
// Find the number of
// digits in number n
int length = (int)floor(log10(n) + 1);
// Append 9
for(int i = 1; i <= n - (2 * length); i++)
{
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = to_string(n) + result + to_string(n);
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
string findNumberS(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
string result = "";
// Find the number of
// digits in number n
int length = (int)floor(log10(n) + 1);
for (int i = 1; i <= n - (2 * length); i++)
{
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = to_string(n) + result + to_string(n);
// Return the smallest number
return result;
}
// Driver code
int main()
{
// Given number
int N = 3;
// Function call
cout << "Smallest Number = " << findNumberS(N) << endl;
cout << "Largest Number = " << findNumberL(N);
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to find n digit
// largest number starting
// and ending with n
static String findNumberL(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
String result = "";
// Find the number of
// digits in number n
int length
= (int)Math.floor(
Math.log10(n) + 1);
// Append 9
for (int i = 1;
i <= n - (2 * length); i++) {
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = Integer.toString(n)
+ result
+ Integer.toString(n);
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
static String findNumberS(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
String result = "";
// Find the number of
// digits in number n
int length
= (int)Math.floor(
Math.log10(n) + 1);
for (int i = 1; i <= n - (2 * length); i++) {
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = Integer.toString(n)
+ result
+ Integer.toString(n);
// Return the smallest number
return result;
}
// Driver Code
public static void main(String[] args)
{
// Given Number
int N = 3;
// Function Call
System.out.println(
"Smallest Number = "
+ findNumberS(N));
System.out.print(
"Largest Number = "
+ findNumberL(N));
}
}Python3
# Python3 program for the
# above approach
import math
# Function to find n digit
# largest number starting
#and ending with n
def findNumberL(n):
# Corner Case when n = 1
if (n == 1):
return "1"
# Result will store the
# n - 2*length(n) digit
# largest number
result = ""
# Find the number of
# digits in number n
length = math.floor(math.log10(n) + 1)
# Append 9
for i in range(1, n - (2 *
length) + 1):
result += '9'
# To make it largest n digit
# number starting and ending
# with n, we just need to
# append n at start and end
result = (str(n) + result +
str(n))
# Return the largest number
return result
# Function to find n digit
# smallest number starting
# and ending with n
def findNumberS(n):
# Corner Case when n = 1
if (n == 1):
return "1"
# Result will store the
# n - 2*length(n) digit
# smallest number
result = ""
# Find the number of
# digits in number n
length = math.floor(math.log10(n) + 1)
for i in range(1, n -
(2 * length) + 1):
result += '0'
# To make it smallest n digit
# number starting and ending
# with n, we just need to
# append n at start and end
result = (str(n) + result +
str(n))
# Return the smallest number
return result
# Driver Code
if __name__ == "__main__":
# Given Number
N = 3
# Function Call
print("Smallest Number = " + findNumberS(N))
print("Largest Number = "+ findNumberL(N))
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
class GFG{
// Function to find n digit
// largest number starting
// and ending with n
static String findNumberL(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// largest number
String result = "";
// Find the number of
// digits in number n
int length = (int)Math.Floor(
Math.Log10(n) + 1);
// Append 9
for(int i = 1;
i <= n - (2 * length); i++)
{
result += '9';
}
// To make it largest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = n.ToString() + result +
n.ToString();
// Return the largest number
return result;
}
// Function to find n digit
// smallest number starting
// and ending with n
static String findNumberS(int n)
{
// Corner Case when n = 1
if (n == 1)
return "1";
// Result will store the
// n - 2*length(n) digit
// smallest number
String result = "";
// Find the number of
// digits in number n
int length = (int)Math.Floor(
Math.Log10(n) + 1);
for (int i = 1;
i <= n - (2 * length); i++)
{
result += '0';
}
// To make it smallest n digit
// number starting and ending
// with n, we just need to
// append n at start and end
result = n.ToString() + result +
n.ToString();
// Return the smallest number
return result;
}
// Driver Code
public static void Main(String[] args)
{
// Given number
int N = 3;
// Function call
Console.WriteLine("Smallest Number = " +
findNumberS(N));
Console.Write("Largest Number = " +
findNumberL(N));
}
}
// This code is contributed by Amit Katiyar输出：
Smallest Number = 303
Largest Number = 393时间复杂度： O(N)