重复给定字符串的子字符串所需的次数 |第 2 组（递归）

给定字符串str，任务是重复字符串X的每个子字符串的次数，其中X是由原始字符串中的子字符串之后出现的连续数字组成的数字。例如，如果str = “g1e2ks1”，那么结果字符串将是“geeks”。

例子：

Input: str = “2a10bd3”
Output: aaaaaaaaaabdbdbd
Explanation: First digit “2” is unnecessary as there is no valid substring before it. Character “a” will be repeated 10 times and then “bd” will be repeated thrice.

Input: str = “g1e2ks1for1g1e2ks1”
Output: geeksforgeeks

这里讨论了解决这个问题的迭代方法。本文重点介绍解决给定问题的递归方法。

方法：递归地逐个字符字符遍历字符串。遍历时，将字符保留在单独的字符串中，将表示重复次数的数字保留在整数中。对于找到的每一组数字，根据找到的整数将存储的字符串的出现附加到最终结果字符串中，并递归调用剩余的字符串。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Stores the final string
string res = "";
 
// Recursive function to generate
// the required string
void decode(string s, int i, string t, int x)
{
    // If complete string has
    // been traversed
    if (i == s.length()) {
 
        // Append string t, s times
        for (int i = 0; i < x; i++)
            res = res + t;
        return;
    }
 
    // If current character
    // is an integer
    if (isdigit(s[i]) && !t.empty()) {
        x = x * 10 + (s[i] - '0');
    }
 
    // If current character
    // in an alphabet
    if (!isdigit(s[i])) {
        if (!t.empty() && x > 0) {
 
            // Append t, x times
            for (int i = 0; i < x; i++)
                res = res + t;
 
            // Update the value
            // of t and x
            t = "";
            x = 0;
        }
        t = t + s[i];
    }
 
    // Recursive call for the
    // remaining string
    decode(s, i + 1, t, x);
}
 
// Function to convert the given
// string into desired form
string decodeString(string s)
{
    // Recursive Call
    decode(s, 0, "", 0);
 
    // Return Answer
    return res;
}
 
// Driven Program
int main()
{
    string str = "g1e2k1s1for1g1e2ks1";
    cout << decodeString(str);
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
 
class GFG
{
 
  // Stores the final String
  static String res = "";
 
  // Recursive function to generate
  // the required String
  static void decode(String s, int i, String t, int x)
  {
 
    // If complete String has
    // been traversed
    if (i == s.length()) {
 
      // Append String t, s times
      for (int j = 0; j < x; j++)
        res = res + t;
      return;
    }
 
    // If current character
    // is an integer
    if (Character.isDigit(s.charAt(i)) && !t.isEmpty()) {
      x = x * 10 + (s.charAt(i) - '0');
    }
 
    // If current character
    // in an alphabet
    if (!Character.isDigit(s.charAt(i))) {
      if (!t.isEmpty() && x > 0) {
 
        // Append t, x times
        for (int j = 0; j < x; j++)
          res = res + t;
 
        // Update the value
        // of t and x
        t = "";
        x = 0;
      }
      t = t + s.charAt(i);
    }
 
    // Recursive call for the
    // remaining String
    decode(s, i + 1, t, x);
  }
 
  // Function to convert the given
  // String into desired form
  static String decodeString(String s)
  {
    // Recursive Call
    decode(s, 0, "", 0);
 
    // Return Answer
    return res;
  }
 
  // Driven Program
  public static void main(String[] args)
  {
    String str = "g1e2k1s1for1g1e2ks1";
    System.out.print(decodeString(str));
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program of the above approach
 
# Stores the final String
res = "";
 
# Recursive function to generate
# the required String
def decode(s, i, t, x):
    global res;
     
    # If complete String has
    # been traversed
    if (i == len(s)):
 
        # Append String t, s times
        for j in range(x):
            res = res + t;
        return;
 
    # If current character
    # is an integer
    if (s[i].isdigit() and len(t)!=0):
        x = x * 10 + (ord(s[i]) - ord('0'));
 
    # If current character
    # in an alphabet
    if (s[i].isdigit()==False):
        if (len(t)!=0 and x > 0):
 
            # Append t, x times
            for j in range(x):
                res = res + t;
 
            # Update the value
            # of t and x
            t = "";
            x = 0;
 
        t = t + s[i];
 
    # Recursive call for the
    # remaining String
    decode(s, i + 1, t, x);
 
# Function to convert the given
# String into desired form
def decodeString(s):
    # Recursive Call
    decode(s, 0, "", 0);
 
    # Return Answer
    return res;
 
# Driven Program
if __name__ == '__main__':
    str = "g1e2k1s1for1g1e2ks1";
    print(decodeString(str));
 
# This code is contributed by 29AjayKumar

C#

// C# program of the above approach
using System;
 
public class GFG
{
 
  // Stores the readonly String
  static String res = "";
 
  // Recursive function to generate
  // the required String
  static void decode(String s, int i, String t, int x)
  {
 
    // If complete String has
    // been traversed
    if (i == s.Length) {
 
      // Append String t, s times
      for (int j = 0; j < x; j++)
        res = res + t;
      return;
    }
 
    // If current character
    // is an integer
    if (char.IsDigit(s[i]) && t.Length!=0) {
      x = x * 10 + (s[i] - '0');
    }
 
    // If current character
    // in an alphabet
    if (!char.IsDigit(s[i])) {
      if (t.Length!=0 && x > 0) {
 
        // Append t, x times
        for (int j = 0; j < x; j++)
          res = res + t;
 
        // Update the value
        // of t and x
        t = "";
        x = 0;
      }
      t = t + s[i];
    }
 
    // Recursive call for the
    // remaining String
    decode(s, i + 1, t, x);
  }
 
  // Function to convert the given
  // String into desired form
  static String decodeString(String s)
  {
    // Recursive Call
    decode(s, 0, "", 0);
 
    // Return Answer
    return res;
  }
 
  // Driven Program
  public static void Main(String[] args)
  {
    String str = "g1e2k1s1for1g1e2ks1";
    Console.Write(decodeString(str));
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

       // JavaScript code for the above approach
 
       // Stores the final string
       var res = "";
 
       // Recursive function to generate
       // the required string
       function decode(s, i, t, x)
       {
        
           // If complete string has
           // been traversed
           if (i == s.length) {
 
               // Append string t, s times
               for (let i = 0; i < x; i++)
                   res = res + t;
               return;
           }
 
           // If current character
           // is an integer
           if (Number.isInteger(parseInt(s[i])) && t.length != 0) {
               x = x * 10 + (s[i].charCodeAt(0) - '0'.charCodeAt(0));
           }
 
           // If current character
           // in an alphabet
           if (!Number.isInteger(parseInt(s[i]))) {
               if (!t.length == 0 && x > 0) {
 
                   // Append t, x times
                   for (let i = 0; i < x; i++)
                       res = res + t;
 
                   // Update the value
                   // of t and x
                   t = "";
                   x = 0;
               }
               t = t + s[i];
           }
 
           // Recursive call for the
           // remaining string
           decode(s, i + 1, t, x);
       }
 
       // Function to convert the given
       // string into desired form
       function decodeString(s)
       {
        
           // Recursive Call
           decode(s, 0, "", 0);
 
           // Return Answer
           return res;
       }
 
       // Driven Program
       let str = "g1e2k1s1for1g1e2ks1";
       document.write(decodeString(str));
 
 // This code is contributed by Potta Lokesh
   

输出

geeksforgeeks

时间复杂度： O(M)，其中 M 是给定字符串中存在的所有整数的总和
辅助空间： O(N)