检查给定的二进制字符串跟在给定的条件之后

给定二进制字符串str ，任务是检查给定的字符串是否符合以下条件：

• 字符串以‘1’开头。
• 每个‘1’后跟空字符串( “” )、 ‘1’或“00” 。
• 每个“00”后跟空字符串( “” )， ‘1’ 。

如果给定的字符串符合上述条件，则打印“有效字符串”，否则打印“无效字符串” 。
例子：

Input: str = “1000” 
Output: False 
Explanation: 
The given string starts with “1” and has “00” followed by the “1” which is not the given criteria. 
Hence, the given string is “Invalid String”.
Input: str = “1111” 
Output: True 
Explanation: 
The given string starts with 1 and has 1 followed by all the 1’s. 
Hence, the given string is “Valid String”. 
 

方法：想法是使用递归。以下是步骤：

1. 检查^第0^个字符是否为“1” 。如果它不是‘1’ ，则返回 false，因为该字符串不符合条件 1。
2. 要检查满足第二个条件的字符串，请使用 C++ 中的 substr()函数从第一个索引开始递归调用字符串。
3. 要检查满足第三个条件的字符串，首先需要检查字符串长度是否大于 2。如果是，则检查第一个和第二个索引处是否存在“0” 。如果是，则递归调用从第 3 个索引开始的字符串。
4. 在任何递归调用中，如果字符串为空，那么我们已经遍历了满足所有给定条件的完整字符串并打印“有效字符串” 。
5. 在任何递归调用中，如果给定条件不满足，则停止该递归并打印“无效字符串” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if the
// string follows rules or not
bool checkrules(string s)
{
    if (s.length() == 0)
        return true;
 
    // Check for the first condition
    if (s[0] != '1')
        return false;
 
    // Check for the third condition
    if (s.length() > 2) {
        if (s[1] == '0' && s[2] == '0')
            return checkrules(s.substr(3));
    }
 
    // Check for the second condition
    return checkrules(s.substr(1));
}
 
// Driver Code
int main()
{
    // Given String str
    string str = "1111";
 
    // Function Call
    if (checkrules(str)) {
        cout << "Valid String";
    }
    else {
        cout << "Invalid String";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if the
// String follows rules or not
static boolean checkrules(String s)
{
    if (s.length() == 0)
        return true;
 
    // Check for the first condition
    if (s.charAt(0) != '1')
        return false;
 
    // Check for the third condition
    if (s.length() > 2)
    {
        if (s.charAt(1) == '0' &&
            s.charAt(2) == '0')
            return checkrules(s.substring(3));
    }
 
    // Check for the second condition
    return checkrules(s.substring(1));
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String str
    String str = "1111";
 
    // Function call
    if (checkrules(str))
    {
        System.out.print("Valid String");
    }
    else
    {
        System.out.print("Invalid String");
    }
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach
 
# Function to check if the
# string follows rules or not
def checkrules(s):
     
    if len(s) == 0:
        return True
         
    # Check for the first condition
    if s[0] != '1':
        return False
         
    # Check for the third condition
    if len(s) > 2:
        if s[1] == '0' and s[2] == '0':
            return checkrules(s[3:])
             
    # Check for the second condition
    return checkrules(s[1:])
     
# Driver code
if __name__ == '__main__':
     
    # Given string
    s = '1111'
     
    # Function call
    if checkrules(s):
        print('valid string')
    else:
        print('invalid string')
         
# This code is contributed by virusbuddah_

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if the
// String follows rules or not
static bool checkrules(String s)
{
    if (s.Length == 0)
        return true;
 
    // Check for the first condition
    if (s[0] != '1')
        return false;
 
    // Check for the third condition
    if (s.Length > 2)
    {
        if (s[1] == '0' &&
            s[2] == '0')
            return checkrules(s.Substring(3));
    }
 
    // Check for the second condition
    return checkrules(s.Substring(1));
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String str
    String str = "1111";
 
    // Function call
    if (checkrules(str))
    {
        Console.Write("Valid String");
    }
    else
    {
        Console.Write("Invalid String");
    }
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript program for the above approach
 
// Function to check if the
// string follows rules or not
function checkrules(s)
{
    if (s.length == 0)
        return true;
 
    // Check for the first condition
    if (s[0] != '1')
        return false;
 
    // Check for the third condition
    if (s.length > 2) {
        if (s[1] == '0' && s[2] == '0')
            return checkrules(s.substring(3));
    }
 
    // Check for the second condition
    return checkrules(s.substring(1));
}
 
// Driver Code
// Given String str
var str = "1111";
// Function Call
if (checkrules(str)) {
    document.write( "Valid String");
}
else {
    document.write( "Invalid String");
}
 

输出：

Valid String

时间复杂度： O(N) ，其中 N 是字符串的长度。
辅助空间： O(1) 。