计算没有连续 1 的二进制字符串的数量：设置 2

给定一个正整数N ，任务是计算所有可能的长度为 N 的不同二进制字符串，以便没有连续的 1。

例子：

Input: N = 5 
Output: 5 
Explanation: 
The non-negative integers <= 5 with their corresponding binary representations are: 
0 : 0 
1 : 1 
2 : 10 
3 : 11 
4 : 100 
5 : 101 
Among them, only 3 has two consecutive 1’s. Therefore required count = 5

Input: N = 12 
Output: 8 
 

动态编程方法：本文已经讨论了动态编程方法。

方法：在本文中，讨论了一种使用 digit-dp 概念的方法。

• 类似于 digit-dp 问题，这里创建了一个3 维表来存储计算值。假设N < 2 ³¹ – 1，并且每个数字的范围只有2(0或1)。因此，该表的尺寸为32 x 2 x 2 。
• 构建表后，给定的数字被转换为二进制字符串。
• 然后，对数字进行迭代。对于每次迭代：
  1. 检查前一个数字是 0 还是 1。
  2. 如果是 0，则当前数字可以是 0 或 1。
  3. 但是如果前一个数字是 1，那么当前数字必须是 0，因为在二进制表示中我们不能有两个连续的 1。
• 现在，表格就像数字 dp 问题一样被填满。

下面是上述方法的实现

C++

// C++ program to count number of
// binary strings without consecutive 1’s
 
#include 
using namespace std;
 
// Table to store the solution of
// every sub problem
int memo[32][2][2];
 
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
int dp(int pos, int fl, int pr, string& bin)
{
    // Base case
    if (pos == bin.length())
        return 1;
 
    // Check if this subproblem
    // has already been solved
    if (memo[pos][fl][pr] != -1)
        return memo[pos][fl][pr];
 
    int val = 0;
 
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin[pos] == '0')
        val = val + dp(pos + 1, fl, 0, bin);
 
    // Here flag will be 1 for the
    // next recursive call
    else if (bin[pos] == '1')
        val = val + dp(pos + 1, 1, 0, bin);
 
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
 
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
 
        // If the digit at current position is 1
        else if (bin[pos] == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
 
    // Storing the solution to this subproblem
    return memo[pos][fl][pr] = val;
}
 
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
int findIntegers(int num)
{
    // Convert N to binary form
    string bin;
 
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2)
            bin += "1";
        else
            bin += "0";
        num /= 2;
    }
    reverse(bin.begin(), bin.end());
 
    // Initialising the table with -1.
    memset(memo, -1, sizeof(memo));
 
    // Calling the function
    return dp(0, 0, 0, bin);
}
 
// Driver code
int main()
{
    int N = 12;
    cout << findIntegers(N);
 
    return 0;
}

Java

// Java program to count number of
// binary Strings without consecutive 1’s
class GFG{
  
// Table to store the solution of
// every sub problem
static int [][][]memo = new int[32][2][2];
  
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
static int dp(int pos, int fl, int pr, String bin)
{
    // Base case
    if (pos == bin.length())
        return 1;
  
    // Check if this subproblem
    // has already been solved
    if (memo[pos][fl][pr] != -1)
        return memo[pos][fl][pr];
  
    int val = 0;
  
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin.charAt(pos) == '0')
        val = val + dp(pos + 1, fl, 0, bin);
  
    // Here flag will be 1 for the
    // next recursive call
    else if (bin.charAt(pos) == '1')
        val = val + dp(pos + 1, 1, 0, bin);
  
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
  
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
  
        // If the digit at current position is 1
        else if (bin.charAt(pos) == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
  
    // Storing the solution to this subproblem
    return memo[pos][fl][pr] = val;
}
  
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
static int findIntegers(int num)
{
    // Convert N to binary form
    String bin = "";
  
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2 == 1)
            bin += "1";
        else
            bin += "0";
        num /= 2;
    }
    bin = reverse(bin);
  
    // Initialising the table with -1.
    for(int i = 0; i < 32; i++){
        for(int j = 0; j < 2; j++){
            for(int l = 0; l < 2; l++)
                memo[i][j][l] = -1;
        }
    }
  
    // Calling the function
    return dp(0, 0, 0, bin);
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 12;
    System.out.print(findIntegers(N));
  
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python program to count number of
# binary strings without consecutive 1’s
 
 
 
# Table to store the solution of
# every sub problem
memo=[[[-1 for i in range(2)] for j in range(2)] for k in range(32)]
 
# Function to fill the table
''' Here,
pos: keeps track of current position.
f1: is the flag to check if current
        number is less than N or not.
pr: represents the previous digit
'''
def dp(pos,fl,pr,bin):
    # Base case
    if (pos == len(bin)):
        return 1;
 
    # Check if this subproblem
    # has already been solved
    if (memo[pos][fl][pr] != -1):
        return memo[pos][fl][pr];
 
    val = 0
 
    # Placing 0 at the current position
    # as it does not violate the condition
    if (bin[pos] == '0'):
        val = val + dp(pos + 1, fl, 0, bin)
 
    # Here flag will be 1 for the
    # next recursive call
    elif (bin[pos] == '1'):
        val = val + dp(pos + 1, 1, 0, bin)
 
    # Placing 1 at this position only if
    # the previously inserted number is 0
    if (pr == 0):
 
        # If the number is smaller than N
        if (fl == 1):
            val += dp(pos + 1, fl, 1, bin)
 
        # If the digit at current position is 1
        elif (bin[pos] == '1'):
            val += dp(pos + 1, fl, 1, bin)
         
    # Storing the solution to this subproblem
    memo[pos][fl][pr] = val
    return val
 
# Function to find the number of integers
# less than or equal to N with no
# consecutive 1’s in binary representation
def findIntegers(num):
    # Convert N to binary form
    bin=""
 
    # Loop to convert N
    # from Decimal to binary
    while (num > 0):
        if (num % 2):
            bin += "1"
        else:
            bin += "0"
        num //= 2
     
    bin=bin[::-1];
 
     
 
    # Calling the function
    return dp(0, 0, 0, bin)
 
# Driver code
if __name__ == "__main__":
 
    N = 12
    print(findIntegers(N))

C#

// C# program to count number of
// binary Strings without consecutive 1’s
using System;
 
public class GFG{
   
// Table to store the solution of
// every sub problem
static int [,,]memo = new int[32,2,2];
   
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
static int dp(int pos, int fl, int pr, String bin)
{
    // Base case
    if (pos == bin.Length)
        return 1;
   
    // Check if this subproblem
    // has already been solved
    if (memo[pos,fl,pr] != -1)
        return memo[pos,fl,pr];
   
    int val = 0;
   
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin[pos] == '0')
        val = val + dp(pos + 1, fl, 0, bin);
   
    // Here flag will be 1 for the
    // next recursive call
    else if (bin[pos] == '1')
        val = val + dp(pos + 1, 1, 0, bin);
   
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
   
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
   
        // If the digit at current position is 1
        else if (bin[pos] == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
   
    // Storing the solution to this subproblem
    return memo[pos,fl,pr] = val;
}
   
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
static int findints(int num)
{
    // Convert N to binary form
    String bin = "";
   
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2 == 1)
            bin += "1";
        else
            bin += "0";
        num /= 2;
    }
    bin = reverse(bin);
   
    // Initialising the table with -1.
    for(int i = 0; i < 32; i++){
        for(int j = 0; j < 2; j++){
            for(int l = 0; l < 2; l++)
                memo[i,j,l] = -1;
        }
    }
   
    // Calling the function
    return dp(0, 0, 0, bin);
}
static String reverse(String input) {
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 12;
    Console.Write(findints(N));
   
}
}
  
// This code contributed by Rajput-Ji

Javascript

 
// JavaScript program to count number of
// binary strings without consecutive 1’s
 
// Table to store the solution of
// every sub problem
var memo = Array.from(Array(32), ()=>Array(2));
 
for(var i =0;i<32;i++)
{
    for(var j =0; j<2; j++)
    {
        memo[i][j] = new Array(2).fill(-1);
    }
}
 
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
function dp(pos, fl, pr, bin)
{
    // Base case
    if (pos == bin.length)
        return 1;
 
    // Check if this subproblem
    // has already been solved
    if (memo[pos][fl][pr] != -1)
        return memo[pos][fl][pr];
 
    var val = 0;
 
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin[pos] == '0')
        val = val + dp(pos + 1, fl, 0, bin);
 
    // Here flag will be 1 for the
    // next recursive call
    else if (bin[pos] == '1')
        val = val + dp(pos + 1, 1, 0, bin);
 
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
 
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
 
        // If the digit at current position is 1
        else if (bin[pos] == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
 
    // Storing the solution to this subproblem
    return memo[pos][fl][pr] = val;
}
 
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
function findIntegers(num)
{
    // Convert N to binary form
    var bin = "";
 
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2)
            bin += "1";
        else
            bin += "0";
        num =parseInt(num/2);
    }
 
    bin = bin.split('').reverse().join('')
 
 
    // Calling the function
    return dp(0, 0, 0, bin);
}
 
// Driver code
var N = 12;
document.write( findIntegers(N));
 

输出：

8

时间复杂度： O(L * log(N))

• O(log(N)) 将数字从十进制转换为二进制。
• O(L) 来填表，其中 L 是二进制形式的长度。

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