2的出现次数(从0到n的数字)

在从0到n的所有数字中，将2s的数量计为数字。
例子 ：

Input : 22
Output : 6
Explanation: Total 2s that appear as digit
             from 0 to 22 are (2, 12, 20,
             21, 22);

Input : 100
Output : 20
Explanation: total 2's comes between 0 to 100 
are (2, 12, 20, 21, 22..29, 32, 42, 52, 62, 72,
82, 92);

一个简单的蛮力解决方案是遍历从0到n的所有数字。对于访问的每个数字，请计算其中的2。最后返回总数。
以下是该想法的实现。

C++

// C++ program to count 2s from 0 to n
#include 
using namespace std;
 
// Counts the number of '2'
// digits in a single number
int number0f2s(int n)
{
    int count = 0;
    while (n > 0)
    {
        if (n % 10 == 2)
            count++;
 
        n = n / 10;
    }
    return count;
}
 
// Counts the number of '2'
// digits between 0 and n
int numberOf2sinRange(int n)
{
    // Initialize result
    int count = 0 ;
 
    // Count 2's in every number
    // from 2 to n
    for (int i = 2; i <= n; i++)
        count += number0f2s(i);
 
    return count;
}
 
// Driver Code
int main()
{
    cout << numberOf2sinRange(22);
    cout << endl;
    cout << numberOf2sinRange(100);
    return 0;
}

Java

// Java program to count 2s from 0 to n
 
class GFG
{
     
// Counts the number of '2'
// digits in a single number
static int number0f2s(int n)
{
    int count = 0;
    while (n > 0)
    {
        if (n % 10 == 2)
        count++;
 
        n = n / 10;
    }
    return count;
}
 
// Counts the number of '2'
// digits between 0 and n
static int numberOf2sinRange(int n)
{
     
    // Initialize result
    int count = 0;
 
    // Count 2's in every number
    // from 2 to n
    for (int i = 2; i <= n; i++)
    count += number0f2s(i);
 
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    System.out.print(numberOf2sinRange(22));
    System.out.println();
    System.out.print(numberOf2sinRange(100));
}
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to count
# 2s from 0 to n
 
# Counts the number of
# '2' digits in a
# single number
def number0f2s(n):
 
    count = 0
    while (n > 0):
     
        if (n % 10 == 2):
            count = count + 1
 
        n = n // 10
     
    return count
 
 
# Counts the number of '2'
# digits between 0 and n
def numberOf2sinRange(n):
 
    # Initialize result
    count = 0
 
    # Count 2's in every number
    # from 2 to n
    for i in range(2, n + 1):
        count = count + number0f2s(i)
 
    return count
 
# Driver code
 
print(numberOf2sinRange(22))
print(numberOf2sinRange(100))
 
# This code is contributed
# by Anant Agarwal.

C#

// C# program to count 2s from 0 to n
using System;
 
class GFG
{
     
// Counts the number of '2' digits
// in a single number
static int number0f2s(int n)
{
    int count = 0;
    while (n > 0)
    {
        if (n % 10 == 2)
            count++;
 
    n = n / 10;
    }
    return count;
}
 
// Counts the number of '2' digits
// between 0 and n
static int numberOf2sinRange(int n)
{
     
    // Initialize result
    int count = 0;
 
    // Count 2's in every number
    // from 2 to n
    for (int i = 2; i <= n; i++)
    count += number0f2s(i);
 
    return count;
}
 
// Driver code
public static void Main()
{
    Console.Write(numberOf2sinRange(22));
    Console.WriteLine();
    Console.Write(numberOf2sinRange(100));
}
}
 
// This code is contributed by nitin mittal

PHP

 0)
    {
        if ($n % 10 == 2)
            $count++;
 
        $n = $n / 10;
    }
    return $count;
}
 
// Counts the number of '2'
// digits between 0 and n
function numberOf2sinRange($n)
{
     
    // Initialize result
    $count = 0 ;
 
    // Count 2's in every number
    // from 2 to n
    for ($i = 2; $i <= $n; $i++)
        $count += number0f2s($i);
 
    return $count;
}
 
// Driver Code
    echo (numberOf2sinRange(22));
    echo "\n";
    echo numberOf2sinRange(100);
 
// This code is contributed by
// nitin mittal.
?>

Javascript

 
    // JavaScript program to count 2s from 0 to n
     
    // Counts the number of '2'
    // digits in a single number
    function number0f2s(n)
    {
        let count = 0;
        while (n > 0)
        {
            if (n % 10 == 2)
                count++;
 
            n = parseInt(n / 10, 10);
        }
        return count;
    }
 
    // Counts the number of '2'
    // digits between 0 and n
    function numberOf2sinRange(n)
    {
        // Initialize result
        let count = 0 ;
 
        // Count 2's in every number
        // from 2 to n
        for (let i = 2; i <= n; i++)
            count += number0f2s(i);
 
        return count;
    }
     
    document.write(numberOf2sinRange(22) + "</br>");
    document.write(numberOf2sinRange(100));
     

输出：

6
20

以下是此方法在Python的替代实现。

C++

// Write C++ code here
#include 
using namespace std;
int numberOf2sinRange(int n)
{
    string s = "";
    for(int i = 0; i < n + 1; i++)
        s += to_string(i);       
    int count = 0;
    for(int i = 0; i < s.length(); i++)
    {
        if(s[i] == '2')
        {
            count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int n = 30;
    cout << numberOf2sinRange(n);
    return 0;
}
 
// This code is contributed by divyeshrabadiya07.

Java

// Java code for above implementation
class GFG
{
  static int numberOf2sinRange(int n)
  {
    String s = "";
    for(int i = 0; i < n + 1; i++)
      s += String.valueOf(i);      
    int count = 0;
    for(int i = 0; i < s.length(); i++)
    {
      if(s.charAt(i) == '2')
      {
        count++;
      }
    }
    return count;
  }
 
  // Driver code
  public static void main(String[] args) {
    int n = 30;
    System.out.println(numberOf2sinRange(n));
  }
}
 
// This code is contributed by divyesh072019

Python3

# Write Python3 code here
def numberOf2sinRange(n):
    s = ""
    for i in range(0,n+1):
        s+=str(i)
    return(list(s).count('2'))
 
# Driver code
n = 30
print(numberOf2sinRange(n))

C＃

// C# code for above implementation
using System;
class GFG
{
  static int numberOf2sinRange(int n)
  {
    string s = "";
    for(int i = 0; i < n + 1; i++)
      s += i + "";  
    int count = 0;
    for(int i = 0; i < s.Length; i++)
    {
      if(s[i] == '2')
      {
        count++;
      }
    }
    return count;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int n = 30;
    Console.Write(numberOf2sinRange(n));
  }
}
 
// This code is contributed by rutvik_56.

输出：

13

改进的解决方案
想法是逐位查看问题。描绘一个数字序列：

0  1  2  3  4  5  6  7  8  9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
......
110 111 112 113 114 115 116 117 118 119 

我们知道大约有十分之一的时间，最后一位数字将是2，因为它以十个数字的任何顺序出现一次。实际上，任何数字大约是十分之一的2。
我们说“大致”是因为存在(非常常见的)边界条件。例如，在1和100之间，10级的数字是时的2恰好^1/10。然而，1和37之间，10级的数字是一个2远远超过^1/10的时间。
我们可以通过分别查看以下三种情况来算出比率的确切数字：数字2。
个案数字<2
考虑值x = 61523，索引d = 3处的数字(此处从右开始考虑索引，最右端的索引为0)。我们观察到x [d] =1。在2000 – 2999、12000 – 12999、22000 – 22999、32000 32999、42000 – 42999和52000 – 52999范围的第三位有2s。因此总共有6000 2在第3位数字中。这与我们只计算第三个数字中所有1到60000之间的2相同。
换句话说，我们可以向下舍入到最接近的10 ^{d + 1} ，然后除以10，以计算第d位数字的2位数。

if x[d) < 2: count2sinRangeAtDigit(x, d) =
  Compute y = round down to nearest 10d+1 
  return y/10 

案例数字> 2
现在，让我们看一下x的第d个数字(从右开始)大于2(x [d]> 2)的情况。我们可以应用几乎完全相同的逻辑来观察，第三位数字中0到63525范围内的2位数与0到70000范围内的位数相同。因此，我们四舍五入而不是四舍五入。

if x[d) > 2: count2sinRangeAtDigit(x, d) =
  Compute y = round down to nearest 10d+1 
  return y / 10 

案例数= 2
最终的情况可能是最棘手的，但是它遵循了先前的逻辑。考虑x = 62523和d =3。我们知道从2s到之前存在相同的范围(即2000 – 2999、12000 – 12999，…，52000 – 52999)。在最后的部分数字(从62000 – 62523)的第3位数字中出现几个?好吧，那应该很容易。只是524(62000，62001，…，62523)。

if x[d] = 2: count2sinRangeAtDigit(x, d) = 
   Compute y = round down to nearest 10d+1 
   Compute z = right side of x (i.e., x%  10d)
   return y/10 + z + 1

现在，我们所需要做的就是遍历数字中的每个数字。实施此代码相当简单。
以下是该想法的实现。

C++

// C++ program to count 2s from 0 to n
#include 
using namespace std;
 
// Counts the number of 2s
// in a number at d-th digit
int count2sinRangeAtDigit(int number, int d)
{
    int powerOf10 = (int)pow(10, d);
    int nextPowerOf10 = powerOf10 * 10;
    int right = number % powerOf10;
 
    int roundDown = number - number % nextPowerOf10;
    int roundup = roundDown + nextPowerOf10;
 
    int digit = (number / powerOf10) % 10;
 
    // if the digit in spot digit is
    if (digit < 2)
        return roundDown / 10;
 
    if (digit == 2)
        return roundDown / 10 + right + 1;
 
    return roundup / 10;
}
 
// Counts the number of '2' digits between 0 and n
int numberOf2sinRange(int number)
{
    // Convert integer to String
    // to find its length
    stringstream convert;
    convert << number;
    string s = convert.str();
    int len = s.length();
 
    // Traverse every digit and
    // count for every digit
    int count = 0;
    for (int digit = 0; digit < len; digit++)
        count += count2sinRangeAtDigit(number, digit);
 
    return count;
}
 
// Driver Code
int main()
{
    cout << numberOf2sinRange(22) << endl;
    cout << numberOf2sinRange(100);
    return 0;
}

Java

// Java program to count 2s from 0 to n
class GFG
{
 
    // Counts the number of 2s
    // in a number at d-th digit
    static int count2sinRangeAtDigit(int number, int d)
    {
        int powerOf10 = (int) Math.pow(10, d);
        int nextPowerOf10 = powerOf10 * 10;
        int right = number % powerOf10;
 
        int roundDown = number - number % nextPowerOf10;
        int roundup = roundDown + nextPowerOf10;
 
        int digit = (number / powerOf10) % 10;
 
        // if the digit in spot digit is
        if (digit < 2)
        {
            return roundDown / 10;
        }
 
        if (digit == 2)
        {
            return roundDown / 10 + right + 1;
        }
        return roundup / 10;
    }
 
    // Counts the number of '2' digits between 0 and n
    static int numberOf2sinRange(int number)
    {
        // Convert integer to String
        // to find its length
        String convert;
        convert = String.valueOf(number);
        String s = convert;
        int len = s.length();
 
        // Traverse every digit and
        // count for every digit
        int count = 0;
        for (int digit = 0; digit < len; digit++)
        {
            count += count2sinRangeAtDigit(number, digit);
        }
 
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        System.out.println(numberOf2sinRange(22));
        System.out.println(numberOf2sinRange(100));
    }
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to count 2s from 0 to n
 
# Counts the number of 2s in a
# number at d-th digit
def count2sinRangeAtDigit(number, d):
 
    powerOf10 = int(pow(10, d));
    nextPowerOf10 = powerOf10 * 10;
    right = number % powerOf10;
 
    roundDown = number - number % nextPowerOf10;
    roundup = roundDown + nextPowerOf10;
 
    digit = (number // powerOf10) % 10;
 
    # if the digit in spot digit is
    if (digit < 2):
        return roundDown // 10;
 
    if (digit == 2):
        return roundDown // 10 + right + 1;
 
    return roundup // 10;
 
# Counts the number of '2' digits
# between 0 and n
def numberOf2sinRange(number):
     
    # Convert integer to String
    # to find its length
    s = str(number);
    len1 = len(s);
 
    # Traverse every digit and
    # count for every digit
    count = 0;
    for digit in range(len1):
        count += count2sinRangeAtDigit(number,
                                       digit);
 
    return count;
 
# Driver Code
print(numberOf2sinRange(22));
print(numberOf2sinRange(100));
 
# This code is contributed by mits

C＃

// C# program to count 2s from 0 to n
using System;
 
class GFG
{
 
// Counts the number of 2s
// in a number at d-th digit
static int count2sinRangeAtDigit(int number,
                                 int d)
{
    int powerOf10 = (int) Math.Pow(10, d);
    int nextPowerOf10 = powerOf10 * 10;
    int right = number % powerOf10;
 
    int roundDown = number - number % nextPowerOf10;
    int roundup = roundDown + nextPowerOf10;
 
    int digit = (number / powerOf10) % 10;
 
    // if the digit in spot digit is
    if (digit < 2)
    {
        return roundDown / 10;
    }
 
    if (digit == 2)
    {
        return roundDown / 10 + right + 1;
    }
    return roundup / 10;
}
 
// Counts the number of '2' digits
// between 0 and n
static int numberOf2sinRange(int number)
{
    // Convert integer to String
    // to find its length
    string convert;
    convert = number.ToString();
    string s = convert;
    int len = s.Length;
 
    // Traverse every digit and
    // count for every digit
    int count = 0;
    for (int digit = 0; digit < len; digit++)
    {
        count += count2sinRangeAtDigit(number,
                                       digit);
    }
 
    return count;
}
 
// Driver Code
public static void Main()
{
    Console.WriteLine(numberOf2sinRange(22));
    Console.WriteLine(numberOf2sinRange(100));
}
}
 
// This code is contributed by mits

输出：

6
20