生成长度为n的所有二进制字符串，其中子字符串“01”恰好出现两次

2021年4月22日14:57:42 发表评论 1,462 次浏览

本文概述

给定一个整数ñ, 任务是生成所有可能的长度为二进制的字符串ñ其中包含" 01"作为子字符串恰好两次。

例子：

输入：N = 4
输出：0101" 0101"是唯一一个长度为4的二进制字符串, 其中包含的" 01"是子字符串的两倍。
输入：N = 5
输出：00101 01001 01010 01011 01101 10101

方法：这个问题可以解决使用回溯。为了生成二进制字符串, 我们实现了一个函数, 该函数一次生成每个位, 更新二进制字符串的状态(当前长度, 模式出现的次数)。然后递归调用该函数, 并根据二进制字符串的当前状态, 该函数将决定如何生成下一位或打印出二进制字符串(如果满足问题要求)。

对于这个问题，回溯策略看起来就像我们生成了一棵二叉树，每个节点的值可以是0或1。

例如，当N = 4时，树看起来如下:

生成长度为n的所有二进制字符串，其中子字符串“ 01”恰好出现两次1

下面是上述方法的实现：

C ++

//C++ implementation of the approach
#include <iostream>
#include <stdlib.h>
using namespace std;
  
//Utility function to print the given binary string
void printBinStr( int * str, int len)
{
     for ( int i = 0; i <len; i++) {
         cout <<str[i];
     }
     cout <<endl;
}
  
//This function will be called recursively
//to generate the next bit for given
//binary string according to its current state
void generateBinStr( int * str, int len, int currlen, int occur, int nextbit)
{
  
     //Base-case: if the generated binary string
     //meets the required length and the pattern "01"
     //appears twice
     if (currlen == len) {
  
         //nextbit needs to be  0 because each time
         //we call the function recursively, //we call 2 times for 2 cases:
         //next bit is 0 or 1
         //The is to assure that the binary
         //string is printed one time only
         if (occur == 2 && nextbit == 0)
             printBinStr(str, len);
         return ;
     }
  
     //Generate the next bit for str
     //and call recursive
     if (currlen == 0) {
  
         //Assign first bit
         str[0] = nextbit;
  
         //The next generated bit will wither be 0 or 1
         generateBinStr(str, len, currlen + 1, occur, 0);
         generateBinStr(str, len, currlen + 1, occur, 1);
     }
     else {
  
         //If pattern "01" occurrence is <2
         if (occur <2) {
  
             //Set next bit
             str[currlen] = nextbit;
  
             //If pattern "01" appears then
             //increase the occurrence of pattern
             if (str[currlen - 1] == 0 && nextbit == 1) {
                 occur += 1;
             }
             generateBinStr(str, len, currlen + 1, occur, 0);
             generateBinStr(str, len, currlen + 1, occur, 1);
  
             //Else pattern "01" occurrence equals 2
         }
         else {
  
             //If previous bit is 0 then next bit cannot be 1
             if (str[currlen - 1] == 0 && nextbit == 1) {
                 return ;
  
                 //Otherwise
             }
             else {
                 str[currlen] = nextbit;
                 generateBinStr(str, len, currlen + 1, occur, 0);
                 generateBinStr(str, len, currlen + 1, occur, 1);
             }
         }
     }
}
  
//Driver code
int main()
{
  
     int n = 5;
  
     //Length of the resulting strings
     //must be at least 4
     if (n <4)
         cout <<-1;
     else {
         int * str = new int [n];
  
         //Generate all binary strings of length n
         //with sub-string "01" appearing twice
         generateBinStr(str, n, 0, 0, 0);
         generateBinStr(str, n, 0, 0, 1);
     }
  
     return 0;
}

Java

//Java implementation of the above approach 
class GFG 
{
  
     //Utility function to print the given binary string
     static void printBinStr( int [] str, int len) 
     {
         for ( int i = 0 ; i <len; i++)
         {
             System.out.print(str[i]);
         }
         System.out.println();
     }
  
     //This function will be called recursively
     //to generate the next bit for given
     //binary string according to its current state
     static void generateBinStr( int [] str, int len, int currlen, int occur, int nextbit) 
     {
  
         //Base-case: if the generated binary string
         //meets the required length and the pattern "01"
         //appears twice
         if (currlen == len) 
         {
  
             //nextbit needs to be 0 because each time
             //we call the function recursively, //we call 2 times for 2 cases:
             //next bit is 0 or 1
             //The is to assure that the binary
             //string is printed one time only
             if (occur == 2 && nextbit == 0 )
             {
                 printBinStr(str, len);
             }
             return ;
         }
  
         //Generate the next bit for str
         //and call recursive
         if (currlen == 0 ) 
         {
  
             //Assign first bit
             str[ 0 ] = nextbit;
  
             //The next generated bit will wither be 0 or 1
             generateBinStr(str, len, currlen + 1 , occur, 0 );
             generateBinStr(str, len, currlen + 1 , occur, 1 );
         } else //If pattern "01" occurrence is <2
         if (occur <2 ) 
         {
  
             //Set next bit
             str[currlen] = nextbit;
  
             //If pattern "01" appears then
             //increase the occurrence of pattern
             if (str[currlen - 1 ] == 0 && nextbit == 1 ) 
             {
                 occur += 1 ;
             }
             generateBinStr(str, len, currlen + 1 , occur, 0 );
             generateBinStr(str, len, currlen + 1 , occur, 1 );
  
             //Else pattern "01" occurrence equals 2
         } else //If previous bit is 0 then next bit cannot be 1
         if (str[currlen - 1 ] == 0 && nextbit == 1 ) 
         {
             return ;
  
             //Otherwise
         } 
         else 
         {
             str[currlen] = nextbit;
             generateBinStr(str, len, currlen + 1 , occur, 0 );
             generateBinStr(str, len, currlen + 1 , occur, 1 );
         }
     }
  
     //Driver code
     public static void main(String[] args) 
     {
         int n = 5 ;
  
         //Length of the resulting strings
         //must be at least 4
         if (n <4 ) 
         {
             System.out.print(- 1 );
         } 
         else
         {
             int [] str = new int [n];
  
             //Generate all binary strings of length n
             //with sub-string "01" appearing twice
             generateBinStr(str, n, 0 , 0 , 0 );
             generateBinStr(str, n, 0 , 0 , 1 );
         }
     }
}
  
//This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
  
# Utility function to print the 
# given binary string
def printBinStr(string, length):
  
     for i in range ( 0 , length): 
         print (string[i], end = "")
      
     print ()
  
# This function will be called recursively
# to generate the next bit for given
# binary string according to its current state
def generateBinStr(string, length, currlen, occur, nextbit):
  
     # Base-case: if the generated binary
     # string meets the required length and
     # the pattern "01" appears twice
     if currlen = = length:
  
         # nextbit needs to be 0 because each
         # time we call the function recursively, # we call 2 times for 2 cases:
         # next bit is 0 or 1
         # The is to assure that the binary
         # string is printed one time only
         if occur = = 2 and nextbit = = 0 :
             printBinStr(string, length)
         return
  
     # Generate the next bit for
     # str and call recursive
     if currlen = = 0 : 
  
         # Assign first bit
         string[ 0 ] = nextbit
  
         # The next generated bit will 
         # either be 0 or 1
         generateBinStr(string, length, currlen + 1 , occur, 0 )
         generateBinStr(string, length, currlen + 1 , occur, 1 )
      
     else :
  
         # If pattern "01" occurrence is <2
         if occur <2 : 
  
             # Set next bit
             string[currlen] = nextbit
  
             # If pattern "01" appears then
             # increase the occurrence of pattern
             if string[currlen - 1 ] = = 0 and nextbit = = 1 :
                 occur + = 1
              
             generateBinStr(string, length, currlen + 1 , occur, 0 )
             generateBinStr(string, length, currlen + 1 , occur, 1 )
  
             # Else pattern "01" occurrence equals 2
          
         else :
  
             # If previous bit is 0 then next bit cannot be 1
             if string[currlen - 1 ] = = 0 and nextbit = = 1 : 
                 return
  
                 # Otherwise
              
             else :
                 string[currlen] = nextbit
                 generateBinStr(string, length, currlen + 1 , occur, 0 )
                 generateBinStr(string, length, currlen + 1 , occur, 1 )
  
# Driver code
if __name__ = = "__main__" :
  
     n = 5
  
     # Length of the resulting strings
     # must be at least 4
     if n <4 :
         print ( - 1 )
     else :
         string = [ None ] * n
  
         # Generate all binary strings of length n
         # with sub-string "01" appearing twice
         generateBinStr(string, n, 0 , 0 , 0 )
         generateBinStr(string, n, 0 , 0 , 1 )
      
# This code is contributed by Rituraj Jain

C#

//C# implementation of the above approach 
using System;
      
class GFG 
{
  
//Utility function to print the given binary string
static void printBinStr( int [] str, int len) 
{
     for ( int i = 0; i <len; i++)
     {
         Console.Write(str[i]);
     }
     Console.Write( "\n" );
}
  
//This function will be called recursively
//to generate the next bit for given
//binary string according to its current state
static void generateBinStr( int [] str, int len, int currlen, int occur, int nextbit) 
{
  
     //Base-case: if the generated binary string
     //meets the required length and the pattern "01"
     //appears twice
     if (currlen == len) 
     {
  
         //nextbit needs to be 0 because each time
         //we call the function recursively, //we call 2 times for 2 cases:
         //next bit is 0 or 1
         //The is to assure that the binary
         //string is printed one time only
         if (occur == 2 && nextbit == 0)
         {
             printBinStr(str, len);
         }
         return ;
     }
  
     //Generate the next bit for str
     //and call recursive
     if (currlen == 0) 
     {
  
         //Assign first bit
         str[0] = nextbit;
  
         //The next generated bit will wither be 0 or 1
         generateBinStr(str, len, currlen + 1, occur, 0);
         generateBinStr(str, len, currlen + 1, occur, 1);
     } else //If pattern "01" occurrence is <2
     if (occur <2) 
     {
  
         //Set next bit
         str[currlen] = nextbit;
  
         //If pattern "01" appears then
         //increase the occurrence of pattern
         if (str[currlen - 1] == 0 && nextbit == 1) 
         {
             occur += 1;
         }
         generateBinStr(str, len, currlen + 1, occur, 0);
         generateBinStr(str, len, currlen + 1, occur, 1);
  
         //Else pattern "01" occurrence equals 2
     } else //If previous bit is 0 then next bit cannot be 1
     if (str[currlen - 1] == 0 && nextbit == 1) 
     {
         return ;
  
         //Otherwise
     } 
     else
     {
         str[currlen] = nextbit;
         generateBinStr(str, len, currlen + 1, occur, 0);
         generateBinStr(str, len, currlen + 1, occur, 1);
     }
}
  
//Driver code
public static void Main(String[] args) 
{
     int n = 5;
  
     //Length of the resulting strings
     //must be at least 4
     if (n <4) 
     {
         Console.Write(-1);
     } 
     else
     {
         int [] str = new int [n];
  
         //Generate all binary strings of length n
         //with sub-string "01" appearing twice
         generateBinStr(str, n, 0, 0, 0);
         generateBinStr(str, n, 0, 0, 1);
     }
}
}
  
//This code is contributed by Princi Singh

输出如下：

00101
01001
01010
01011
01101
10101
木子山

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