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📜  删除字符串所有出现的字符|递归方法

📅  最后修改于: 2021-06-26 16:14:07             🧑  作者: Mango

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:

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:
    • 基本案例: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(除了基本情况和上述条件之外),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)

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