递归程序用给定字符串中的3.14替换所有出现的pi

本文概述

C ++
Java
Python3
C#
python
C#

这个任务是写一个递归函数来替换所有出现在给定字符串中的pi为3.14，并打印修改后的字符串。
例子：

输入：str ="pippppiiiipi"
输出：3.14ppp3.14iii3.14
输入：str ="pip"
输出：3.14p
输入：str ="xpix"
输出：x3.14x

我们在这里讨论了一个迭代函数
方法：

如果字符串中只有一个字符或该字符串为空, 则中断递归调用
否则, 你自己保留字符串的第一个字符, 然后将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"并且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 < bits/stdc++.h> 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] == '\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

//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; } }

输出如下

3.14ppp3.14iii3.14

另一种方法：
一种简单的递归方法, 用" 3.14"替换给定函数中的所有pi。首先声明函数, 我们不需要任何辅助函数。

如果字符串为空或字符串的长度为1, 则返回字符串。
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：
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# 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

【递归程序用给定字符串中的3.14替换所有出现的pi】输出如下：