最大偶数不大于N

给定一个数字N，我们需要编写一个程序以查找不大于N且具有全数字的最大数字。
例子：

Input: N = 23
Output: 22 
Explanation: 22 is the largest number not 
greater then N which has all digits even. 

Input: N = 236
Output: 228 
Explanation: 228 is the largest number not 
greater than N which has all digits even. 

幼稚的方法：幼稚的方法是从N迭代到0，并找到第一个具有所有数字偶数的数字。
下面是上述方法的实现：

C++

// CPP program to print the largest
// integer not greater than N with all even digits
#include 
using namespace std;
 
// function to check if all digits
// are even of a given number
int checkDigits(int n)
{
    // iterate for all digits
    while (n) {
        if ((n % 10) % 2) // if digit is odd
            return 0;
 
        n /= 10;
    }
 
    // all digits are even
    return 1;
}
 
// function to return the largest number
// with all digits even
int largestNumber(int n)
{
    // iterate till we find a
    // number with all digits even
    for (int i = n;; i--)
        if (checkDigits(i))
            return i;
}
 
// Driver Code
int main()
{
    int N = 23;   
    cout << largestNumber(N);
    return 0;
}

Java

// Java program to print the largest
// integer not greater than N with
// all even digits
import java .io.*;
 
public class GFG {
     
// function to check if all digits
// are even of a given number
static int checkDigits(int n)
{
     
    // iterate for all digits
    while (n > 0)
    {
         
        // if digit is odd
        if (((n % 10) % 2) > 0)
            return 0;
 
        n /= 10;
    }
 
    // all digits are even
    return 1;
}
 
// function to return the largest
// number with all digits even
static int largestNumber(int n)
{
     
    // iterate till we find a
    // number with all digits even
    for (int i = n;; i--)
        if (checkDigits(i) > 0)
            return i;
}
 
    // Driver Code
    static public void main (String[] args)
    {
        int N = 23;
        System.out.println(largestNumber(N));
    }
}
 
// This code is contributed by vt_m.

Python3

# Python3 program to print the largest
# integer not greater than N with
# all even digits
 
# function to check if all digits
# are even of a given number
def checkDigits(n):
 
     
    # iterate for all digits
    while (n>0):
        # if digit is odd
        if ((n % 10) % 2):
            return False;
 
        n =int(n/10);
 
    # all digits are even
    return True;
 
# function to return the
# largest number with
# all digits even
def largestNumber(n):
     
    # Iterate till we find a
    # number with all digits even
    for i in range(n,-1,-1):
        if (checkDigits(i)):
            return i;
 
# Driver Code
N = 23;
print(largestNumber(N));
 
# This code is contributed by mits

C#

// C# program to print the largest
// integer not greater than N with
// all even digits
using System;
 
public class GFG {
     
// function to check if all digits
// are even of a given number
static int checkDigits(int n)
{
     
    // iterate for all digits
    while (n > 0)
    {
         
        // if digit is odd
        if (((n % 10) % 2) > 0)
            return 0;
 
        n /= 10;
    }
 
    // all digits are even
    return 1;
}
 
// function to return the largest
// number with all digits even
static int largestNumber(int n)
{
     
    // iterate till we find a
    // number with all digits even
    for (int i = n;; i--)
        if (checkDigits(i) > 0)
            return i;
}
 
    // Driver Code
    static public void Main ()
    {
        int N = 23;
        Console.WriteLine(largestNumber(N));
    }
}
 
// This code is contributed by aunj_67.

Javascript

 
// javascript program to prvar the largest
// integer not greater than N with
// all even digits
 
// function to check if all digits
// are even of a given number
function checkDigits(n)
{
     
    // iterate for all digits
    while (n > 0)
    {
         
        // if digit is odd
        if (((n % 10) % 2) > 0)
            return 0;
 
        n = parseInt(n/ 10);
    }
 
    // all digits are even
    return 1;
}
 
// function to return the largest
// number with all digits even
function largestNumber(n)
{
     
    // iterate till we find a
    // number with all digits even
    for (i = n;; i--)
        if (checkDigits(i) > 0)
            return i;
}
 
// Driver Code
var N = 23;
document.write(largestNumber(N));
 
// This code is contributed by shikhasingrajput

输出：

22

时间复杂度： O(N)
高效的方法：我们可以通过将N中的第一个奇数位减1，然后用最大的偶数(即8)替换该奇数位右边的所有数字来获得所需的数字。例如，如果N = 24578，则X =24488。在某些情况下，此方法可以创建前导0，在这种情况下，我们可以简单地删除前导0。例如，如果N = 1334，则X =0888。因此，我们的答案将是X =888。如果N中没有奇数位，则N就是数字本身。
下面是上述方法的实现：

C++

// CPP program to print the largest
// integer not greater than N with all even digits
#include 
using namespace std;
 
// function to return the largest number
// with all digits even
int largestNumber(int n)
{
    string s = "";
    int duplicate = n;
 
    // convert the number to a string for
    // easy operations
    while (n) {
        s = char(n % 10 + 48) + s;
        n /= 10;
    }
 
    int index = -1;
 
    // find first odd digit
    for (int i = 0; i < s.length(); i++) {
        if ((s[i] - '0') % 2 & 1) {
            index = i;
            break;
        }
    }
 
    // if no digit, then N is the answer
    if (index == -1)
        return duplicate;
 
    int num = 0;
 
    // till first odd digit, add all even numbers
    for (int i = 0; i < index; i++)
        num = num * 10 + (s[i] - '0');
 
    // decrease 1 from the odd digit
    num = num * 10 + (s[index] - '0' - 1);
 
    // add 0 in the rest of the digits
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10 + 8;
 
    return num;
}
 
// Driver Code
int main()
{
    int N = 24578;
 
    cout << largestNumber(N);
 
    return 0;
}

Java

// Java program to print the largest
// integer not greater than N with all even digits
class GFG
{
     
// function to return the largest number
// with all digits even
static int largestNumber(int n)
{
    String s = "";
    int duplicate = n;
 
    // convert the number to a string for
    // easy operations
    while (n > 0)
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }
 
    int index = -1;
 
    // find first odd digit
    for (int i = 0; i < s.length(); i++)
    {
        if ((((int)(s.charAt(i) - '0') % 2) & 1) > 0)
        {
            index = i;
            break;
        }
    }
 
    // if no digit, then N is the answer
    if (index == -1)
        return duplicate;
 
    int num = 0;
 
    // till first odd digit, add all even numbers
    for (int i = 0; i < index; i++)
        num = num * 10 + (int)(s.charAt(i) - '0');
 
    // decrease 1 from the odd digit
    num = num * 10 + ((int)s.charAt(index) - (int)('0') - 1);
 
    // add 0 in the rest of the digits
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10 + 8;
 
    return num;
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 24578;
 
    System.out.println(largestNumber(N));
}
}
 
// This code is contributed by mits

Python3

# Python3 program to print the largest
# integer not greater than N with
# all even digits
import math as mt
 
# function to return the largest
# number with all digits even
def largestNumber(n):
 
    s = ""
    duplicate = n
 
    # convert the number to a string
    # for easy operations
    while (n > 0):
        s = chr(n % 10 + 48) + s
        n = n // 10
     
    index = -1
 
    # find first odd digit
    for i in range(len(s)):
        if ((ord(s[i]) - ord('0')) % 2 & 1):
            index = i
            break
         
    # if no digit, then N is the answer
    if (index == -1):
        return duplicate
 
    num = 0
 
    # till first odd digit, add all
    # even numbers
    for i in range(index):
        num = num * 10 + (ord(s[i]) - ord('0'))
 
    # decrease 1 from the odd digit
    num = num * 10 + (ord(s[index]) -  
                      ord('0') - 1)
 
    # add 0 in the rest of the digits
    for i in range(index+1,len(s)):
        num = num * 10 + 8
 
    return num
 
# Driver Code
N = 24578
 
print(largestNumber(N))
 
# This code is contributed
# by Mohit kumar 29
    

C＃

// C# program to print the largest
// integer not greater than N with all even digits
using System;
 
class GFG
{
     
// function to return the largest number
// with all digits even
static int largestNumber(int n)
{
    string s = "";
    int duplicate = n;
 
    // convert the number to a string for
    // easy operations
    while (n > 0)
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }
 
    int index = -1;
 
    // find first odd digit
    for (int i = 0; i < s.Length; i++)
    {
        if ((((int)(s[i] - '0') % 2) & 1) > 0)
        {
            index = i;
            break;
        }
    }
 
    // if no digit, then N is the answer
    if (index == -1)
        return duplicate;
 
    int num = 0;
 
    // till first odd digit, add all even numbers
    for (int i = 0; i < index; i++)
        num = num * 10 + (int)(s[i] - '0');
 
    // decrease 1 from the odd digit
    num = num * 10 + ((int)s[index] - (int)('0') - 1);
 
    // add 0 in the rest of the digits
    for (int i = index + 1; i < s.Length; i++)
        num = num * 10 + 8;
 
    return num;
}
 
// Driver Code
static void Main()
{
    int N = 24578;
 
    Console.WriteLine(largestNumber(N));
}
}
 
// This code is contributed by mits

Java脚本

// javascript program to prvar the largest
// integer not greater than N with all even digits
 
     
// function to return the largest number
// with all digits even
function largestNumber(n)
{
    var s = "";
    var duplicate = n;
 
    // convert the number to a string for
    // easy operations
    while (n > 0)
    {
        s = String.fromCharCode(n % 10 + 48) + s;
        n = parseInt(n/10);
    }
 
    var index = -1;
 
    // find first odd digit
    for (i = 0; i < s.length; i++)
    {
        if ((((s.charAt(i).charCodeAt(0) - '0'.charCodeAt(0)) % 2) & 1) > 0)
        {
            index = i;
            break;
        }
    }
 
    // if no digit, then N is the answer
    if (index == -1)
        return duplicate;
 
    var num = 0;
 
    // till first odd digit, add all even numbers
    for (i = 0; i < index; i++)
        num = num * 10 + (s.charAt(i).charCodeAt(0) - '0'.charCodeAt(0));
 
    // decrease 1 from the odd digit
    num = num * 10 + (s.charAt(index).charCodeAt(0) - ('0').charCodeAt(0) - 1);
 
    // add 0 in the rest of the digits
    for (i = index + 1; i < s.length; i++)
        num = num * 10 + 8;
 
    return num;
}
 
// Driver Code
var N = 24578;
 
document.write(largestNumber(N));
 
// This code contributed by shikhasingrajput

输出：

24488

时间复杂度： O(M)，其中M是位数