递归程序，用给定字符串的3.14替换所有出现的pi

给定字符串str的大小N。任务是编写一个递归函数，用给定字符串的3.14替换所有出现的pi并打印修改后的字符串。

例子：

Input : str = “pippppiiiipi” 
Output : 3.14ppp3.14iii3.14

Input : str = “pip” 
Output : 3.14p

Input : str = “xpix” 
Output : x3.14x

我们在这里讨论了迭代函数
方法 ：

• 如果字符串只有一个字符或该字符串为空，则中断递归调用
• 否则，保持与自己的字符串的第一个字符和字符串的其余部分传递给递归。
  • 如果第一个字符不是“ p”，则只需将该字符放在递归答案的前面
  • 否则，如果第一个字符为“ p”，而传递给递归的零件的第一个字符为“ i”，则将“ pi”替换为“ 3.14”

下面是上述方法的实现：

C++

// A recursive C++ program to replace
// all pi in a given string with 3.14
#include 
using namespace std;
 
// Recursive Function to replace all
// occurrences of pi in a given
// with 3.14
void replacePiHelper(char str[], int start)
{
 
    // Base condition
    // if the string is empty
    // or of length one
    if (str[start] == '\0' || str[start + 1] == '\0') {
        return;
    }
 
    // Getting the answer from
    // recursion for the smaller
    // problem
    replacePiHelper(str, start + 1);
 
    // Small calculation part
    // if the first character is 'p'
    // and the first character of the part
    // passed to recursion is 'i' then replace
    // "pi" with "3.14"
    if (str[start] == 'p' && str[start + 1] == 'i') {
 
        // Shifting the characeters to
        // right side to put 3.14 in
        // the character array
        for (int i = strlen(str); i >= start + 2; i--) {
            str[i + 2] = str[i];
        }
 
        // Replacing with "3.14"
        str[start] = '3';
        str[start + 1] = '.';
        str[start + 2] = '1';
        str[start + 3] = '4';
    }
}
 
// Function to replace pi with 3.14
void replacePi(char str[])
{
    replacePiHelper(str, 0);
}
 
// Driver code
int main()
{
    char str[] = "pippppiiiipi";
 
    // Function call
    replacePi(str);
 
    cout << str;
 
    return 0;
}

Java

// A recursive Java program to replace
// all pi in a given string with 3.14
 
class GFG {
 
    // Recursive Function to replace all
    // occurrences of pi in a given
    // with 3.14
    public String replacePi(String str)
    {
        // base condition
        // if the string is empty
        // or of length one
        if (str.length() <= 1) {
            return str;
        }
 
        // if the first character is 'p'
        // and the first character of the part
        // passed to recursion is 'i' then replace
        //"pi" with "3.14"
        if (str.charAt(0) == 'p' && str.length() >= 2
            && str.charAt(1) == 'i') {
            return "3.14" + replacePi(str.substring(2, str.length()));
        }
 
        // if the first character is not 'p'
        // then just put that character in
        // front of the answer which came
        // from recursion
        return str.charAt(0) + replacePi(str.substring(1, str.length()));
    }
 
    // Driver Code
    public static void main(String args[])
    {
        GFG g = new GFG();
        String str = "pippppiiiipi";
        System.out.println(g.replacePi(str));
    }
}

Python3

# A recursive Python3 program to replace
# all pi in a given string with 3.14
 
# Recursive Function to replace all
# occurrences of pi in a given
# with 3.14
def replacePieHelper(string, start):
 
    # Base condition
    # if the string is empty
    # or of length one
    if len(string) < 2 or start == len(string):
        return string
 
    # Getting the answer from
    # recursion for the smaller
    # problem
    replacePieHelper(string, start + 1)
 
    # Small calculation part
    # if the first character is 'p'
    # and the first character of the part
    # passed to recursion is 'i' then replace
    # "pi" with "3.14"
    if(string[start] == 'p' and
       string[start + 1] == 'i'):
 
        # Replacing with "3.14"
        string[start:start + 2] = ['3', '.', '1', '4']
 
# Function to replace pi with 3.14
def replacePi(string):
    replacePieHelper(string, 0)
 
# Driver Code
if __name__ == "__main__":
    string = "pippppiiiipi"
 
    string = list(string)
 
    # Function call
    replacePi(string)
 
    string = ''.join(string)
    print(string)
 
# This code is contributed by
# sanjeev2552

C#

// A recursive C# program to replace
// all pi in a given string with 3.14
 
using System;
class gfg {
    // Recursive Function to replace all
    // occurrences of pi in a given
    // with 3.14
    public String replacePi(String str)
    {
        // base condition
        // if the string is empty
        // or of length one
        if (str.Length <= 1) {
            return str;
        }
 
        // if the first character is 'p'
        // and the first character of the part
        // passed to recursion is 'i' then replace
        //"pi" with "3.14"
        if (str[0] == 'p' && str.Length >= 2
            && str[1] == 'i') {
            return "3.14" + replacePi(str.Substring(2, str.Length - 2));
        }
 
        // if the first character is not 'p'
        // then just put that character in
        // front of the answer which came
        // from recursion
        return str[0] + replacePi(str.Substring(1, str.Length - 1));
    }
}
 
// Driver Code
class geek {
    public static int Main()
    {
        gfg g = new gfg();
        string input = "pippppiiiipi";
        Console.WriteLine(g.replacePi(input));
        return 0;
    }
}

输出

3.14ppp3.14iii3.14

另一种方法：

一种简单的递归方法，用“ 3.14”替换给定函数的所有pi。首先声明函数，我们不需要任何辅助函数。

• 碱的情况下，如果该字符串是空的或字符串的长度为1点返回的字符串。
• 如果字符串的第0个元素和第1个元素是p，并且我们必须处理它们的其余部分，则必须调用递归，它将给出结果。
• 如果不是，那么我们必须从1st调用递归到所有元素，然后将递归结果添加到1st元素并返回它。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// A simple recursive approach
// to replace all pi in a given
// function with "3.14". Firstly
// function is declared we don't
// need any helper function one
// function is enough
string replacePi(string s)
{
 
    // Base case if s is empty
    // or length of s is 1
    // return the s
    if (s.length() == 0 || s.length() == 1)
        return s;
 
    // If the 0th and 1st element
    // of s are p and i we have to
    // handle them for rest we have
    // to call recursion it will
    // give the result
    if (s[0] == 'p' && s[1] == 'i') {
 
        // Smalloutput is a variable
        // used to store recursion result
        string smallOutput = replacePi(s.substr(2));
 
        // And we have to add the recursion
        // result with the first part we
        // handeled and return the answer
        return "3.14" + smallOutput;
    }
    else {
        // If 1st & 2nd element aren't "p" & "i", then keep
        // 1st index as it is & call recursion for rest of
        // the string.
        return s[0] + replacePi(s.substr(1));
    }
}
 
// Driver code
int main()
{
    string s = "pipppiiipi";
 
    // Function call
    string result = replacePi(s);
 
    cout << result << endl;
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program for above approach
class GFG {
 
    // A simple recursive approach
    // to replace all pi in a given
    // function with "3.14". Firstly
    // function is declared we don't
    // need any helper function one
    // function is enough
    public static String replacePi(String s)
    {
 
        // Base case if s is empty
        // or length of s is 1
        // return the s
        if (s.length() == 0 || s.length() == 1)
            return s;
 
        // If the 0th and 1st element
        // of s are p and i we have to
        // handle them for rest we have
        // to call recursion it will
        // give the result
        if (s.charAt(0) == 'p' && s.charAt(1) == 'i') {
 
            // Smalloutput is a variable
            // used to store recursion result
            String smallOutput = replacePi(s.substring(2));
 
            // And we have to add the recursion
            // result with the first part we
            // handeled and return the answer
            return "3.14" + smallOutput;
        }
        else {
 
            // If not then we have to call
            // recursion from 1st to all elements
            // then add recursion result to
            // 1st element and return it
            String smallOutput = replacePi(s.substring(1));
            return s.charAt(0) + smallOutput;
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "pipppiiipi";
 
        // Function call
        String result = replacePi(s);
        System.out.println(result);
    }
}
 
// This code is contributed by divyesh072019

Python

# Python program for above approach
 
# A simple recursive approach
# to replace all
# pi in a given function with "3.14"
# Firstly function is declared we don't
# need any helper function one
# function is enough
 
 
def replacePi(string):
 
    # Base case if string is empty
    # or length of string is 1
    # return the string
    if len(string) == 0 or
                     len(string) == 1:
        return string
     
    # If the 0th and 1st element
    # of string are p
    # and i we have to handle them
    # for rest we have to call
    # recursion it will give the result
    if string[0] == 'p' and string[1] == 'i':
       
        # Smalloutput is a variable
        # used to store recursion result
        smallOutput = replacePi(string[2:])   
         
        # And we have to add the recursion
        # result with the first part we
          # handeled and return the answer
        return "3.14" + smallOutput            
    else:
         
        # If not then we have to call
        # recursion from 1st to all elements
        # then add recursion result to
        # 1st element and return it
        smallOutput = replacePi(string[1:])
        return string[0] + smallOutput
       
# Driver code
if __name__ == "__main__":
 
  string = "pipppiiipi"
 
  # Function call
  result = replacePi(string)
  print result

C＃

// C# program for above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
 
class GFG {
 
    // A simple recursive approach
    // to replace all pi in a given
    // function with "3.14". Firstly
    // function is declared we don't
    // need any helper function one
    // function is enough
    static string replacePi(string s)
    {
 
        // Base case if s is empty
        // or length of s is 1
        // return the s
        if (s.Length == 0 || s.Length == 1)
            return s;
 
        // If the 0th and 1st element
        // of s are p and i we have to
        // handle them for rest we have
        // to call recursion it will
        // give the result
        if (s[0] == 'p' && s[1] == 'i') {
 
            // Smalloutput is a variable
            // used to store recursion result
            string smallOutput = replacePi(s.Substring(2));
 
            // And we have to add the recursion
            // result with the first part we
            // handeled and return the answer
            return "3.14" + smallOutput;
        }
        else {
 
            // If not then we have to call
            // recursion from 1st to all elements
            // then add recursion result to
            // 1st element and return it
            string smallOutput = replacePi(s.Substring(1));
            return s[0] + smallOutput;
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        string s = "pipppiiipi";
 
        // Function call
        string result = replacePi(s);
        Console.Write(result);
    }
}
 
// This code is contributed by rutvik_56

输出：

3.14pp3.14ii3.14