用给定字符串的字符Y 替换所有出现的字符X

给定一个字符串str和两个字符X和Y ，任务是编写一个递归函数，用字符Y替换所有出现的字符X 。

例子：

Input: str = abacd, X = a, Y = x 
Output: xbxcd

Input: str = abacd, X = e, Y = y 
Output: abacd  

迭代方法：这个想法是迭代给定的字符串，如果找到任何字符X则用Y替换该字符。

时间复杂度： O(N)
辅助空间： O(1)

递归方法：这个想法是递归遍历给定的字符串，并将值为X的字符替换为Y 。以下是步骤：

1. 获取字符串str 、字符X和Y 。
2. 递归地从索引0迭代到字符串length 。
   • 基本情况：如果我们到达字符串的末尾，则退出函数。

if(str[0]=='\0')
   return ;

• 递归调用：如果不满足基本情况，则检查第 0 个索引处的字符是否为X，然后用Y替换该字符并递归迭代下一个字符。

if(str[0]==X) 
  str[0] = Y

• 返回语句：在每次递归调用时(基本情况除外)，返回下一次迭代的递归函数。

return recursive_function(str + 1, X, Y)

下面是递归实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to replace all occurrences
// of character c1 with character c2
void replaceCharacter(char input[],
                    char c1, char c2)
{
    // Base Case
    // If the string is empty
    if (input[0] == '\0') {
        return;
    }
 
    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    if (input[0] == c1) {
        input[0] = c2;
    }
 
    // Getting the answer from recursion
    // for the smaller problem
    return replaceCharacter(input + 1,
                            c1, c2);
}
 
// Driver Code
int main()
{
    // Given string
    char str[] = "abacd";
    char c1 = 'a';
    char c2 = 'x';
 
    // Function call
    replaceCharacter(str, c1, c2);
 
    // Print the string
    cout << str;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.io.*;
 
class GFG{
     
// Function to replace all occurrences
// of character c1 with character c2
static String replaceCharacter(String str,
                             char c1, char c2)
{
    // Base Case
    // If the string is empty
    if (str.length() == 1)
    {
        return str;
    }
    char x=str.charAt(0);
    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    if (str.charAt(0) == c1)
    {
        x=c2;
        str = c2+str.substring(1);
    }
 
    // Getting the answer from recursion
    // for the smaller problem
    return x+replaceCharacter(str.substring(1),
                            c1, c2);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given string
    String str = "abacd";
    char c1 = 'a';
    char c2 = 'x';
 
    // Function call
    System.out.println(replaceCharacter(str, c1, c2));
}
}
 
// This code is contributed by cyrus18

Python3

# Python3 program for the above approach
 
# Function to replace all occurrences
# of character c1 with character c2
def replaceCharacter(input, c1, c2):
     
    input = list(str)
     
    # If the character at starting
    # of the given string is equal
    # to c1, replace it with c2
    for i in range(0, len(str)):
        if (input[i] == c1):
            input[i] = c2;
             
        # Print the string
        print(input[i], end = "")
     
# Driver Code
 
# Given string
str = "abacd"
c1 = 'a'
c2 = 'x'
 
# Function call
replaceCharacter(str, c1, c2);
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to replace all occurrences
// of character c1 with character c2
static void replaceCharacter(string str,
                             char c1, char c2)
{
    char[] input = str.ToCharArray();
 
    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    for(int i = 0; i < str.Length; i++)
    {
        if (input[i] == c1)
        {
            input[i] = c2;
        }
     
        // Print the string
        Console.Write(input[i]);
    }
}
 
// Driver Code
public static void Main()
{
     
    // Given string
    string str = "abacd";
    char c1 = 'a';
    char c2 = 'x';
 
    // Function call
    replaceCharacter(str, c1, c2);
}
}
 
// This code is contributed by sanjoy_62

Javascript

// javascript program for the above approach
 
     
// Function to replace all occurrences
// of character c1 with character c2
function replaceCharacter(str,c1, c2)
{
    // Base Case
    // If the string is empty
    if (str.length == 1)
    {
        return str;
    }
    var x=str.charAt(0);
    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    if (str.charAt(0) == c1)
    {
        x=c2;
        str = c2+str.substring(1);
    }
 
    // Getting the answer from recursion
    // for the smaller problem
    return x+replaceCharacter(str.substring(1),
                            c1, c2);
}
 
// Driver Code
//Given string
var str = "abacd";
var c1 = 'a';
var c2 = 'x';
 
// Function call
document.write(replaceCharacter(str, c1, c2));
 
// This code is contributed by 29AjayKumar

输出

xbxcd

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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