小于N的二进制数字计数

给定一个限制N，我们需要找出小于N的二进制数字的数量。二进制数字是那些只包含0和1的数字，因为它们的数字为1、10、101等都是二进制数字。

例子：

Input : N = 200
Output : 7
Count of binary digit number smaller than N is 7, 
enumerated below,
1, 10, 11, 110, 101, 100, 111

解决此问题的一种简单方法是从1循环到N，并检查每个数字是否为二进制数字。如果它是一个二进制数字，请增加此类数字的计数，但是此过程将花费O(N)时间。我们可以做得更好，因为我们知道此类数字的数量将远小于N，因此我们只能对二进制数字进行迭代，并检查生成的数字是否小于N。
在下面的代码中，类似于BFS的方法被实现为仅对二进制数字进行迭代。我们从1开始，每次将(t * 10)和(t * 10 +1)推入队列，其中t是弹出的元素，如果t是二进制数字，则(t * 10)和(t * 10 +1)也会是二进制数字，因此我们仅使用队列来遍历这些数字。当弹出的数字越过N时，我们将停止在队列中推送元素。

C++

// C++ program to count all binary digit
// numbers smaller than N
#include 
using namespace std;
  
//  method returns count of binary digit
//  numbers smaller than N
int countOfBinaryNumberLessThanN(int N)
{
    //  queue to store all intermediate binary
    // digit numbers
    queue q;
  
    //  binary digits start with 1
    q.push(1);
    int cnt = 0;
    int t;
  
    //  loop untill we have element in queue
    while (!q.empty())
    {
        t = q.front();
        q.pop();
  
        //  push next binary digit numbers only if
        // current popped element is N
        if (t <= N)
        {
            cnt++;
  
            // uncomment below line to print actual
            // number in sorted order
            // cout << t << " ";
  
            q.push(t * 10);
            q.push(t * 10 + 1);
        }
    }
  
    return cnt;
}
  
//    Driver code to test above methods
int main()
{
    int N = 200;
    cout << countOfBinaryNumberLessThanN(N);
    return 0;
}

Java

import java.util.LinkedList;
import java.util.Queue;
  
// java program to count all binary digit
// numbers smaller than N
public class GFG {
  
//  method returns count of binary digit
//  numbers smaller than N
    static int countOfBinaryNumberLessThanN(int N) {
        //  queue to store all intermediate binary
        // digit numbers
        Queue q = new LinkedList<>();
  
        //  binary digits start with 1
        q.add(1);
        int cnt = 0;
        int t;
  
        //  loop untill we have element in queue
        while (q.size() > 0) {
            t = q.peek();
            q.remove();
  
            //  push next binary digit numbers only if
            // current popped element is N
            if (t <= N) {
                cnt++;
  
                // uncomment below line to print actual
                // number in sorted order
                // cout << t << " ";
                q.add(t * 10);
                q.add(t * 10 + 1);
            }
        }
  
        return cnt;
    }
  
//    Driver code to test above methods
    static public void main(String[] args) {
        int N = 200;
        System.out.println(countOfBinaryNumberLessThanN(N));
    }
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program to count all binary digit
# numbers smaller than N
from collections import deque
  
# method returns count of binary digit
# numbers smaller than N
def countOfBinaryNumberLessThanN(N):
    # queue to store all intermediate binary
    # digit numbers
    q = deque()
  
    # binary digits start with 1
    q.append(1)
    cnt = 0
  
    # loop untill we have element in queue
    while (q):
        t = q.popleft()
          
        # push next binary digit numbers only if
        # current popped element is N
        if (t <= N):
            cnt = cnt + 1
            # uncomment below line to print actual
            # number in sorted order
            q.append(t * 10)
            q.append(t * 10 + 1)
  
    return cnt
  
# Driver code to test above methods
if __name__=='__main__':
    N = 200
    print(countOfBinaryNumberLessThanN(N))
  
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to count all binary digit 
// numbers smaller than N 
using System;
using System.Collections.Generic;
  
class GFG 
{ 
  
    // method returns count of binary digit 
    // numbers smaller than N 
    static int countOfBinaryNumberLessThanN(int N) 
    {
          
        // queue to store all intermediate binary 
        // digit numbers 
        Queue q = new Queue(); 
  
        // binary digits start with 1 
        q.Enqueue(1); 
        int cnt = 0; 
        int t; 
  
        // loop untill we have element in queue 
        while (q.Count > 0)
        { 
            t = q.Peek(); 
            q.Dequeue(); 
  
            // push next binary digit numbers only if 
            // current popped element is N 
            if (t <= N) 
            { 
                cnt++; 
  
                // uncomment below line to print actual 
                // number in sorted order 
                q.Enqueue(t * 10); 
                q.Enqueue(t * 10 + 1); 
            } 
        } 
  
        return cnt; 
    } 
  
    // Driver code  
    static void Main() 
    { 
        int N = 200; 
        Console.WriteLine(countOfBinaryNumberLessThanN(N)); 
    } 
} 
  
// This code is contributed by mits

输出：

7