总数为1个

给定整数n，计算出现在所有小于或等于n的正整数中的数字1的总数。
例子：

Input : n = 13
Output : 6
Explanation:
Here, no <= 13 containing 1 are 1, 10, 11,
12, 13. So, total 1s are 6.

Input : n = 5
Output : 1
Here, no <= 13 containing 1 are 1.
So, total 1s are 1.

方法1：

1. 从1到n遍历i。
2. 将i转换为字符串，并在每个整数字符串计数’1’。
3. 将每个字符串中的计数“ 1”加到总和上。

下面是上述方法的代码。

C++

// c++ code to count the frequency of 1
// in numbers less than or equal to
// the given number.
#include 
using namespace std;
int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i++) {
        string str = to_string(i);
        countr += count(str.begin(), str.end(), '1');
    }
    return countr;
}
 
// driver function
int main()
{
    int n = 13;
    cout << countDigitOne(n) << endl;
    n = 131;
    cout << countDigitOne(n) << endl;
    n = 159;
    cout << countDigitOne(n) << endl;
    return 0;
}

Java

// Java code to count the frequency of 1
// in numbers less than or equal to
// the given number.
class GFG
{
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i++)
    {
        String str = String.valueOf(i);
        countr += str.split("1", -1) . length - 1;
    }
    return countr;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 13;
    System.out.println(countDigitOne(n));
    n = 131;
    System.out.println(countDigitOne(n));
    n = 159;
    System.out.println(countDigitOne(n));
}
}
 
// This code is contributed by mits

Python3

# Python3 code to count the frequency
# of 1 in numbers less than or equal
# to the given number.
 
def countDigitOne(n):
    countr = 0;
    for i in range(1, n + 1):
        str1 = str(i);
        countr += str1.count("1");
 
    return countr;
 
# Driver Code
n = 13;
print(countDigitOne(n));
 
n = 131;
print(countDigitOne(n));
 
n = 159;
print(countDigitOne(n));
 
# This code is contributed by mits

C#

// C# code to count the frequency of 1
// in numbers less than or equal to
// the given number.
using System;
class GFG
{
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i++)
    {
        string str = i.ToString();
        countr += str.Split("1") . Length - 1;
    }
    return countr;
}
 
// Driver Code
public static void Main()
{
    int n = 13;
    Console.WriteLine(countDigitOne(n));
    n = 131;
    Console.WriteLine(countDigitOne(n));
    n = 159;
    Console.WriteLine(countDigitOne(n));
}
}
 
// This code is contributed by mits

输出：

6
66
96

时间复杂度： O(nlog(n))
方法二：

1. 将Counter初始化为0。
2. 将i从1迭代到n，每次递增10。
3. 在每个(i * 10)间隔之后，将(n /(i * 10))* i添加到计数器。
4. 将min(max(n mod(i * 10)– i + 1，0)，i)添加到计数器。

下面是上述方法的代码。

C++

// c++ code to count the frequency of 1
// in numbers less than or equal to
// the given number.
#include 
using namespace std;
 
// function to count the frequency of 1.
int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1; i <= n; i *= 10) {
        int divider = i * 10;
        countr += (n / divider) * i +
               min(max(n % divider - i + 1, 0), i);
    }
    return countr;
}
 
// driver function
int main()
{
    int n = 13;
    cout << countDigitOne(n) << endl;
    n = 113;
    cout << countDigitOne(n) << endl;
    n = 205;
    cout << countDigitOne(n) << endl;
}

Java

/// Java code to count the
// frequency of 1 in numbers
// less than or equal to
// the given number.
import java.io.*;
 
class GFG
{
     
// function to count
// the frequency of 1.
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1;
             i <= n; i *= 10)
    {
        int divider = i * 10;
        countr += (n / divider) * i +
                Math.min(Math.max(n %
                         divider - i +
                            1, 0), i);
    }
    return countr;
}
 
// Driver Code
public static void main (String[] args)
{
    int n = 13;
    System.out.println(countDigitOne(n));
     
    n = 113;
    System.out.println(countDigitOne(n));
     
    n = 205;
    System.out.println(countDigitOne(n));
}
}
 
// This code is contributed by akt_mit

Python3

# Python3 code to count the
# frequency of 1 in numbers
# less than or equal to
# the given number.
 
# function to count the
# frequency of 1.
def countDigitOne(n):
    countr = 0;
    i = 1;
    while(i <= n):
        divider = i * 10;
        countr += (int(n / divider) * i +
                 min(max(n % divider -i +
                              1, 0), i));
        i *= 10;
     
    return countr;
 
# Driver Code
n = 13;
print(countDigitOne(n));
n = 113;
print(countDigitOne(n));
n = 205;
print(countDigitOne(n));
 
# This code is contributed by mits

C＃

// C# code to count the
// frequency of 1 in numbers
// less than or equal to
// the given number.
using System;
 
class GFG
{
     
// function to count
// the frequency of 1.
static int countDigitOne(int n)
{
    int countr = 0;
    for (int i = 1;
             i <= n; i *= 10)
    {
        int divider = i * 10;
        countr += (n / divider) * i +
                   Math.Min(Math.Max(n % divider -
                                     i + 1, 0), i);
    }
    return countr;
}
 
// Driver Code
public static void Main()
{
    int n = 13;
    Console.WriteLine(countDigitOne(n));
     
    n = 113;
    Console.WriteLine(countDigitOne(n));
     
    n = 205;
    Console.WriteLine(countDigitOne(n));
}
}
 
// This code is contributed by mits

Java脚本

 
// Javascript code to count the frequency of 1
// in numbers less than or equal to
// the given number.
 
// function to count the frequency of 1.
function countDigitOne(n)
{
    var countr = 0;
    for (var i = 1; i <= n; i *= 10) {
        var divider = i * 10;
        countr += parseInt(n / divider) * i +
            Math.min(Math.max(n % divider - i + 1, 0), i);
    }
    return countr;
}
 
// driver function
var n = 13;
document.write(countDigitOne(n)+ "<br>");
n = 113;
document.write(countDigitOne(n)+ "<br>");
n = 205;
document.write(countDigitOne(n)+ "<br>");
 
 

输出：

6
40
141

时间复杂度： O(log(n))