删除字符串所有出现的字符|递归方法

给定字符串str ，任务是编写一个递归程序以删除字符串中所有出现的字符X。

例子：

Input: str = “geeksforgeeks”, c = ‘e’ 
Output: gksforgks

Input: str = “geeksforgeeks”, c = ‘g’ 
Output: eeksforeeks 

迭代方法：此问题的迭代方法可以在本文中找到。
递归方法：以下是步骤：

1. 获取字符串str和字符X为这字符X将被移除。
2. 递归地迭代字符串中的所有字符：
   • 基本情况：如果递归调用的字符串str的长度为0，则从函数返回空字符串。

     if(str.length()==0) {
        return "";
     }
     

   • 递归调用：如果不满足基本条件，则检查索引为0的字符是否为X，然后递归迭代子字符串以删除第一个字符。

     if (str[0] == X) {
             return recursive_function(str.substr(1), X);
     }
     

   • 返回语句：在每次递归调用时(基本情况和上述条件除外)，都为下一次迭代返回递归函数，包括第0个index处的字符。

     return str[0] + recursive_function(str.substr(1), X)
     

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to remove all occurrences
// of a character in the string
string removeCharRecursive(string str,
                           char X)
{
    // Base Case
    if (str.length() == 0) {
        return "";
    }
  
    // Check the first character
    // of the given string
    if (str[0] == X) {
  
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(str.substr(1), X);
    }
  
    // Add the first character of str
    // and string from recursion
    return str[0]
           + removeCharRecursive(str.substr(1), X);
}
  
// Driver Code
int main()
{
    // Given String
    string str = "geeksforgeeks";
  
    // Given character
    char X = 'e';
  
    // Function Call
    str = removeCharRecursive(str, X);
    cout << str;
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
// Function to remove all occurrences
// of a character in the string
static String removeCharRecursive(String str,
                                  char X)
{
      
    // Base Case
    if (str.length() == 0)
    {
        return "";
    }
  
    // Check the first character
    // of the given string
    if (str.charAt(0) == X)
    {
  
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(
               str.substring(1), X);
    }
  
    // Add the first character of str
    // and string from recursion
    return str.charAt(0) +
           removeCharRecursive(
           str.substring(1), X);
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given String
    String str = "geeksforgeeks";
  
    // Given character
    char X = 'e';
  
    // Function call
    str = removeCharRecursive(str, X);
      
    System.out.println(str);
}
}
  
// This code is contributed by jrishabh99

Python3

# Python3 program for the above approach
  
# Function to remove all occurrences
# of a character in the string
def removeCharRecursive(str, X):
      
    # Base Case
    if (len(str) == 0):
        return ""
      
    # Check the first character
    # of the given string
    if (str[0] == X):
  
        # Pass the rest of the string
        # to recursion Function call
        return removeCharRecursive(str[1:], X)
      
    # Add the first character of str
    # and string from recursion
    return str[0] + removeCharRecursive(str[1:], X)
  
# Driver Code
  
# Given String
str = "geeksforgeeks"
  
# Given character
X = 'e'
  
# Function call
str = removeCharRecursive(str, X)
  
print(str)
  
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
  
class GFG{
      
// Function to remove all occurrences
// of a character in the string
static String removeCharRecursive(String str,
                                  char X)
{
      
    // Base Case
    if (str.Length == 0)
    {
        return "";
    }
  
    // Check the first character
    // of the given string
    if (str[0] == X)
    {
  
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(
            str.Substring(1), X);
    }
  
    // Add the first character of str
    // and string from recursion
    return str[0] + removeCharRecursive(
                    str.Substring(1), X);
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given String
    String str = "geeksforgeeks";
  
    // Given character
    char X = 'e';
  
    // Function call
    str = removeCharRecursive(str, X);
      
    Console.WriteLine(str);
}
}
  
// This code is contributed by Amit Katiyar

输出：

gksforgks

时间复杂度： O(N)，其中N是字符串的长度
辅助空间： O(1)