最小奇数个数不少于N

给定数字N，任务是找到不小于N的最小数字，该数字的所有数字均为奇数。

例子：

Input: N = 1345  
Output: 1351
1351 is the smallest number not 
less than N, whose all digits are odd. 

Input: N = 2397 
Output: 3111 
3111 is the smallest number not 
less than N, whose all digits are odd.

幼稚的方法：幼稚的方法是保持从N进行迭代，直到找到一个所有数字都为奇数的数字。

下面是上述方法的实现：

C++

// CPP program to print the smallest
// integer not less than N with all odd digits
#include 
using namespace std;
  
// function to check if all digits
// are odd of a given number
int check_digits(int n)
{
    // iterate for all digits
    while (n) {
        if ((n % 10) % 2 == 0) // if digit is even
            return 0;
  
        n /= 10;
    }
  
    // all digits are odd
    return 1;
}
  
// function to return the smallest number
// with all digits odd
int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i++)
        if (check_digits(i))
            return i;
}
  
// Driver Code
int main()
{
    int N = 2397;
    cout << smallest_number(N);
  
    return 0;
}

Java

// Java program to print the smallest
// integer not less than N with all odd digits
class Geeks {
  
// function to check if all digits
// are odd of a given number
static int check_digits(int n)
{
    // iterate for all digits
    while (n > 0) {
        if ((n % 10) % 2 == 0) // if digit is even
            return 0;
  
        n /= 10;
    }
  
    // all digits are odd
    return 1;
}
  
// function to return the smallest number
// with all digits odd
static int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i++)
        if (check_digits(i) > 0)
            return i;
}
  
// Driver Code
public static void main(String args[])
{
    int N = 2397;
    System.out.println(smallest_number(N));
}
}
  
// This code is contributed by Kirti_Mangal

Python3

# Python 3 program to print the smallest
# integer not less than N with all odd digits
  
# function to check if all digits
# are odd of a given number
def check_digits(n):
      
    # iterate for all digits
    while (n):
          
        # if digit is even
        if ((n % 10) % 2 == 0):
            return 0
  
        n = int(n / 10)
  
    # all digits are odd
    return 1
  
# function to return the smallest 
# number with all digits odd
def smallest_number(n):
      
    # iterate till we find a
    # number with all digits odd
    i = n
    while(1):
        if (check_digits(i)):
            return i
  
        i += 1
  
# Driver Code
if __name__ == '__main__':
    N = 2397
    print(smallest_number(N))
  
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to print the smallest
// integer not less than N with all
// odd digits
using System;
  
class GFG
{
// function to check if all digits
// are odd of a given number
static int check_digits(int n)
{
    // iterate for all digits
    while (n != 0) 
    {
        if ((n % 10) % 2 == 0) // if digit is even
            return 0;
  
        n /= 10;
    }
  
    // all digits are odd
    return 1;
}
  
// function to return the smallest 
// number with all digits odd
static int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i++)
        if (check_digits(i) == 1)
            return i;
}
  
// Driver Code
static void Main()
{
    int N = 2397;
    Console.WriteLine(smallest_number(N));
}
}
  
// This code is contributed by ANKITRAI1

PHP

 1) 
    {
        // if digit is even
        if (($n % 10) % 2 == 0) 
            return 0;
  
        $n = (int)$n / 10;
    }
  
    // all digits are odd
    return 1;
}
  
// function to return the smallest 
// number with all digits odd
function smallest_number( $n)
{
    // iterate till we find a
    // number with all digits odd
    for ($i = $n;; $i++)
        if (check_digits($i))
            return $i;
}
  
// Driver Code
$N = 2397;
echo smallest_number($N);
  
// This code is contributed by ajit
?>

输出：

3111

时间复杂度： O(N)

高效的方法：我们可以通过将N中的第一个偶数数字增加一个，然后用最小的奇数数字(即1)替换该奇数数字右边的所有数字来找到数字。如果N中没有偶数，则N本身就是最小的数字。例如，考虑N =213。将N中的第一个偶数数字递增，即2到3，然后将其右边的所有数字都替换为1。因此，我们所需的数字将为311。

下面是有效方法的实现：

C++

// CPP program to print the smallest
// integer not less than N with all odd digits
#include 
using namespace std;
  
// function to return the smallest number
// with all digits odd
int smallestNumber(int n)
{
    int num = 0;
    string s = "";
  
    int duplicate = n;
    // convert the number to string to
    // perform operations
    while (n) {
        s = char(n % 10 + 48) + s;
        n /= 10;
    }
  
    int index = -1;
  
    // find out the first even number
    for (int i = 0; i < s.length(); i++) {
        int digit = s[i] - '0';
        if ((digit & 1) == 0) {
            index = i;
            break;
        }
    }
  
    // if no even numbers are there, than n is the answer
    if (index == -1)
        return duplicate;
  
    // add all digits till first even
    for (int i = 0; i < index; i++)
        num = num * 10 + (s[i] - '0');
  
    // increase the even digit by 1
    num = num * 10 + (s[index] - '0' + 1);
  
    // add 1 to the right of the even number
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10 + 1;
  
    return num;
}
  
// Driver Code
int main()
{
    int N = 2397;
    cout << smallestNumber(N);
  
    return 0;
}

Java

//Java program to print the smallest
// integer not less than N with all odd digits
  
public class GFG {
  
// function to return the smallest number
// with all digits odd
    static int smallestNumber(int n) {
        int num = 0;
        String s = "";
  
        int duplicate = n;
        // convert the number to string to
        // perform operations
        while (n > 0) {
            s = (char) (n % 10 + 48) + s;
            n /= 10;
        }
  
        int index = -1;
  
        // find out the first even number
        for (int i = 0; i < s.length(); i++) {
            int digit = s.charAt(i) - '0';
            if ((digit & 1) == 0) {
                index = i;
                break;
            }
        }
  
        // if no even numbers are there, than n is the answer
        if (index == -1) {
            return duplicate;
        }
  
        // add all digits till first even
        for (int i = 0; i < index; i++) {
            num = num * 10 + (s.charAt(i) - '0');
        }
  
        // increase the even digit by 1
        num = num * 10 + (s.charAt(index) - '0' + 1);
  
        // add 1 to the right of the even number
        for (int i = index + 1; i < s.length(); i++) {
            num = num * 10 + 1;
        }
  
        return num;
    }
  
// Driver Code
    static public void main(String[] args) {
        int N = 2397;
        System.out.println(smallestNumber(N));
    }
}
  
/*This code is contributed by PrinciRaj1992*/

的Python 3

# Python 3 program to print the smallest
# integer not less than N with all odd digits
  
# function to return the smallest 
# number with all digits odd
def smallestNumber(n):
  
    num = 0
    s = ""
  
    duplicate = n
      
    # convert the number to string to
    # perform operations
    while (n):
        s = chr(n % 10 + 48) + s
        n //= 10
  
    index = -1
  
    # find out the first even number
    for i in range(len( s)):
        digit = ord(s[i]) - ord('0')
        if ((digit & 1) == 0) :
            index = i
            break
  
    # if no even numbers are there,
    # than n is the answer
    if (index == -1):
        return duplicate
  
    # add all digits till first even
    for i in range( index):
        num = num * 10 + (ord(s[i]) - 
                          ord('0'))
  
    # increase the even digit by 1
    num = num * 10 + (ord(s[index]) - 
                      ord('0') + 1)
  
    # add 1 to the right of the
    # even number
    for i in range(index + 1 , len(s)):
        num = num * 10 + 1
  
    return num
  
# Driver Code
if __name__ == "__main__":
      
    N = 2397
    print(smallestNumber(N))
  
# This code is contributed by ita_c

C＃

// C# program to print the smallest 
// integer not less than N with all odd digits 
  
using System;
public class GFG{
  
  
// function to return the smallest number 
// with all digits odd 
    static int smallestNumber(int n) { 
        int num = 0; 
        String s = ""; 
  
        int duplicate = n; 
        // convert the number to string to 
        // perform operations 
        while (n > 0) { 
            s = (char) (n % 10 + 48) + s; 
            n /= 10; 
        } 
  
        int index = -1; 
  
        // find out the first even number 
        for (int i = 0; i < s.Length; i++) { 
            int digit = s[i] - '0'; 
            if ((digit & 1) == 0) { 
                index = i; 
                break; 
            } 
        } 
  
        // if no even numbers are there, than n is the answer 
        if (index == -1) { 
            return duplicate; 
        } 
  
        // add all digits till first even 
        for (int i = 0; i < index; i++) { 
            num = num * 10 + (s[i] - '0'); 
        } 
  
        // increase the even digit by 1 
        num = num * 10 + (s[index] - '0' + 1); 
  
        // add 1 to the right of the even number 
        for (int i = index + 1; i < s.Length; i++) { 
            num = num * 10 + 1; 
        } 
  
        return num; 
    } 
  
// Driver Code 
    static public void Main() { 
        int N = 2397; 
        Console.WriteLine(smallestNumber(N)); 
    } 
} 
  
// This code is contributed by PrinciRaj1992 

输出：

3111

时间复杂度： O(M)，其中M是N中的位数。