计数具有连续1的字符串

给定数字n，请计算n个长度为1的长度字符串的数目。

例子：

Input  : n = 2
Output : 1
There are 4 strings of length 2, the
strings are 00, 01, 10 and 11. Only the 
string 11 has consecutive 1's.

Input  : n = 3
Output : 3
There are 8 strings of length 3, the
strings are 000, 001, 010, 011, 100, 
101, 110 and 111.  The strings with
consecutive 1's are 011, 110 and 111.

Input : n = 5
Output : 19

可以使用动态编程解决不带连续1的字符串计数的相反问题(请参阅此处的解决方案)。我们可以使用该解决方案，并通过以下步骤找到所需的数量。

1. 使用此处讨论的方法计算不带连续1的二进制字符串的数量。将此计数设为c 。
2. 连续的1的所有可能的二进制字符串的计数为2 ^ n，其中n为输入长度。
3. 连续1的二进制字符串总数为2 ^ n – c。

以下是上述步骤的执行。

C++

// C++ program to count all distinct
// binary strings with two consecutive 1's
#include 
using namespace std;
 
// Returns count of n length binary
// strings with consecutive 1's
int countStrings(int n)
{
    // Count binary strings without consecutive 1's.
    // See the approach discussed on be
    // ( http://goo.gl/p8A3sW )
    int a[n], b[n];
    a[0] = b[0] = 1;
    for (int i = 1; i < n; i++)
    {
        a[i] = a[i - 1] + b[i - 1];
        b[i] = a[i - 1];
    }
 
    // Subtract a[n-1]+b[n-1] from 2^n
    return (1 << n) - a[n - 1] - b[n - 1];
}
 
// Driver code
int main()
{
    cout << countStrings(5) << endl;
    return 0;
}

Java

// Java program to count all distinct
// binary strings with two consecutive 1's
 
class GFG {
 
    // Returns count of n length binary
    // strings with consecutive 1's
    static int countStrings(int n)
    {
        // Count binary strings without consecutive 1's.
        // See the approach discussed on be
        // ( http://goo.gl/p8A3sW )
        int a[] = new int[n], b[] = new int[n];
        a[0] = b[0] = 1;
 
        for (int i = 1; i < n; i++) {
            a[i] = a[i - 1] + b[i - 1];
            b[i] = a[i - 1];
        }
 
        // Subtract a[n-1]+b[n-1]
        from 2 ^ n return (1 << n) - a[n - 1] - b[n - 1];
    }
 
    // Driver code
    public static void main(String args[])
    {
        System.out.println(countStrings(5));
    }
}
 
// This code is contributed by Nikita tiwari.

Python 3

# Python 3 program to count all
# distinct binary strings with
# two consecutive 1's
 
 
# Returns count of n length
# binary strings with
# consecutive 1's
def countStrings(n):
 
    # Count binary strings without
    # consecutive 1's.
    # See the approach discussed on be
    # ( http://goo.gl/p8A3sW )
    a = [0] * n
    b = [0] * n
    a[0] = b[0] = 1
    for i in range(1, n):
        a[i] = a[i - 1] + b[i - 1]
        b[i] = a[i - 1]
 
    # Subtract a[n-1]+b[n-1] from 2^n
    return (1 << n) - a[n - 1] - b[n - 1]
 
 
# Driver code
print(countStrings(5))
 
 
# This code is contributed
# by Nikita tiwari.

C#

// program to count all distinct
// binary strings with two
// consecutive 1's
using System;
 
class GFG {
 
    // Returns count of n length
    // binary strings with
    // consecutive 1's
    static int countStrings(int n)
    {
        // Count binary strings without
        // consecutive 1's.
        // See the approach discussed on
        // ( http://goo.gl/p8A3sW )
        int[] a = new int[n];
        int[] b = new int[n];
        a[0] = b[0] = 1;
 
        for (int i = 1; i < n; i++)
        {
            a[i] = a[i - 1] + b[i - 1];
            b[i] = a[i - 1];
        }
 
        // Subtract a[n-1]+b[n-1]
        // from 2^n
        return (1 << n) - a[n - 1] - b[n - 1];
    }
 
    // Driver code
    public static void Main()
    {
        Console.WriteLine(countStrings(5));
    }
}
 
// This code is contributed by Anant Agarwal.

Javascript

 
// JavaScript program to count all distinct
// binary strings with two
// consecutive 1's
 
    // Returns count of n length binary
    // strings with consecutive 1's
    function countStrings(n)
    {
        // Count binary strings without consecutive 1's.
        // See the approach discussed on be
        // ( http://goo.gl/p8A3sW )
        let a = [], b = [];
        a[0] = b[0] = 1;
  
        for (let i = 1; i < n; i++) {
            a[i] = a[i - 1] + b[i - 1];
            b[i] = a[i - 1];
        }
  
        // Subtract a[n-1]+b[n-1]
        // from 2 ^ n
        return (1 << n) - a[n - 1] - b[n - 1];
    }
 
// Driver Code
 
       document.write(countStrings(5));
 

输出

19

优化：
上述解决方案的时间复杂度为O(n)。我们可以优化上述解决方案以在O(Logn)中工作。
如果我们仔细研究不带连续1的字符串计数模式，我们可以观察到，当n> = 1时，计数实际上是第(n + 2)^个斐波那契数。斐波那契数为0、1、1、2， 3，5，8，13，21，34，55，89，141，…。

n = 1, count = 0  = 21 - fib(3) 
n = 2, count = 1  = 22 - fib(4)
n = 3, count = 3  = 23 - fib(5)
n = 4, count = 8  = 24 - fib(6)
n = 5, count = 19 = 25 - fib(7)
................