检查字符串可以拆分为以 N 开头后跟 N 个字符的子字符串

给定一个字符串str ，任务是检查它是否可以拆分为子字符串，以便每个子字符串以一个数值开头，后跟由该数字整数表示的多个字符。

例子：

Input: str = “4g12y6hunter”
Output: Yes
Explanation:
Substrings “4g12y” and “6hunter” satisfy the given condition

Input: str = “31ba2a”
Output: No
Explanation:
The entire string cannot be split into substrings of desired types

方法：

请按照以下步骤解决问题：
• 检查无法拆分时的条件：
• 如果给定的字符串不以数字开头。
• 如果子串开头的整数大于剩余子串中后续字符的总数。
• 如果不满足以上两个条件，答案肯定是可能的。因此递归地找到子串。

下面是上述方法的实现：

C++

// C++ Program to implement the
// above approach
#include 
using namespace std;
  
// Function to check if the given
// can be split into desired
// substrings
bool helper(string& s, int pos)
{
    // Length of the string
    int len = s.size();
  
    if (pos >= len)
        return true;
  
    if (!isdigit(s[pos]))
        return false;
  
    int num = 0;
  
    // Traverse the string
    for (int i = pos; i < len; i++) {
  
        // Extract the digit
        num = num * 10 + s[pos] - '0';
  
        // Check if the extracted number
        // does not exceed the remaining
        // length
        if (i + 1 + num > len)
            return false;
  
        // Check for the remaining
        // string
        if (helper(s, i + 1 + num))
            return true;
    }
  
    // If generating desired
    // substrings is not possible
    return false;
}
  
// Driver Code
int main()
{
    string s = "123abc4db1c";
    if (helper(s, 0))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program to implement the 
// above approach 
import java.util.*;
  
class GFG{
      
// Function to check if the given 
// can be split into desired 
// substrings 
public static boolean helper(String s, int pos) 
{ 
      
    // Length of the string 
    int len = s.length(); 
      
    if (pos >= len) 
        return true; 
    if (!Character.isDigit(s.charAt(pos)))
        return false; 
      
    int num = 0; 
      
    // Traverse the string 
    for(int i = pos; i < len; i++)
    {
          
       // Extract the digit 
       num = num * 10 + s.charAt(pos) - '0'; 
         
       // Check if the extracted number 
       // does not exceed the remaining 
       // length 
       if (i + 1 + num > len) 
           return false; 
         
       // Check for the remaining 
       // string 
       if (helper(s, i + 1 + num)) 
           return true; 
    } 
      
    // If generating desired 
    // substrings is not possible 
    return false; 
} 
  
// Driver code
public static void main (String[] args)
{
    String s = "123abc4db1c"; 
      
    if (helper(s, 0)) 
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to implement the
# above approach
  
# Function to check if the given
# can be split into desired
# substrings
def helper(s, pos):
      
    # Length of the string
    size = len(s)
  
    if(pos >= size):
        return True
  
    if(s[pos].isdigit() == False):
        return False
  
    num = 0
  
    # Traverse the string
    for i in range(pos, size):
          
        # Extract the digit
        num = num * 10 + ord(s[pos]) - 48
  
        # Check if the extracted number
        # does not exceed the remaining
        # length
        if (i + 1 + num > size):
            return False
  
        # Check for the remaining
        # string
        if (helper(s, i + 1 + num)):
            return True
  
    # If generating desired
    # substrings is not possible
    return False
  
# Driver Code
s = "123abc4db1c";
  
if (helper(s, 0)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Sanjit_Prasad

C#

// C# program to implement the 
// above approach 
using System;
  
class GFG{
       
// Function to check if the given 
// can be split into desired 
// substrings 
public static bool helper(String s, int pos) 
{ 
       
    // Length of the string 
    int len = s.Length; 
       
    if (pos >= len) 
        return true; 
    if (!char.IsDigit(s[pos]))
        return false; 
       
    int num = 0; 
       
    // Traverse the string 
    for(int i = pos; i < len; i++)
    {
           
       // Extract the digit 
       num = num * 10 + s[pos] - '0'; 
          
       // Check if the extracted number 
       // does not exceed the remaining 
       // length 
       if (i + 1 + num > len) 
           return false; 
          
       // Check for the remaining 
       // string 
       if (helper(s, i + 1 + num)) 
           return true; 
    } 
       
    // If generating desired 
    // substrings is not possible 
    return false; 
} 
   
// Driver code
public static void Main(String[] args)
{
    String s = "123abc4db1c"; 
       
    if (helper(s, 0)) 
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by amal kumar choubey

输出：

Yes


时间复杂度： O(N)
辅助空间： O(1)

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