使用递归
将字母‘x’的所有出现从字符串 s 移动到末尾
原文:https://www . geesforgeks . org/move-all-occurrency-of-字母-x-from-string-s-to-end-use-recursion/
给定一个字符串 s ，我们的任务是使用递归将所有出现的字母 x 移动到字符串 s 的末尾。 注意:如果给定的字符串中只有字母 x，那么返回字符串不变。
示例:
输入:s = " geekxsforgexxeksxx " 输出: geeksforgeeksxxxxx 解释: 字母‘x’的所有出现都移动到最后。
输入: s = "xxxxx" 输出: xxxxx 解释: 由于给定字符串中只有字母 x，因此输出不变。
方法: 为了解决上面提到的问题，我们可以使用递归。遍历字符串并递归检查当前字符是否等于字符“x”。如果没有，则打印该字符，否则移动到下一个字符，直到达到字符串的长度。
下面是上述方法的实现:
C++
// C++ implementation to Move all occurrence of letter ‘x’
// from the string s to the end using Recursion
#include <bits/stdc++.h>
using namespace std;
// Function to move all 'x' in the end
void moveAtEnd(string s, int i, int l)
{
if (i >= l)
return;
// Store current character
char curr = s[i];
// Check if current character is not 'x'
if (curr != 'x')
cout << curr;
// recursive function call
moveAtEnd(s, i + 1, l);
// Check if current character is 'x'
if (curr == 'x')
cout << curr;
return;
}
// Driver code
int main()
{
string s = "geekxsforgexxeksxx";
int l = s.length();
moveAtEnd(s, 0, l);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to Move all occurrence of letter ‘x’
// from the string s to the end using Recursion
import java.util.*;
class GFG{
// Function to move all 'x' in the end
static void moveAtEnd(String s, int i, int l)
{
if (i >= l)
return;
// Store current character
char curr = s.charAt(i);
// Check if current character is not 'x'
if (curr != 'x')
System.out.print(curr);
// recursive function call
moveAtEnd(s, i + 1, l);
// Check if current character is 'x'
if (curr == 'x')
System.out.print(curr);
return;
}
// Driver code
public static void main(String args[])
{
String s = "geekxsforgexxeksxx";
int l = s.length();
moveAtEnd(s, 0, l);
}
}
// This code is contributed by Code_Mech
Python 3
# Python3 implementation to move all
# occurrences of letter ‘x’ from the
# string s to the end using recursion
# Function to move all 'x' in the end
def moveAtEnd(s, i, l):
if(i >= l):
return
# Store current character
curr = s[i]
# Check if current character
# is not 'x'
if(curr != 'x'):
print(curr, end = "")
# Recursive function call
moveAtEnd(s, i + 1, l)
# Check if current character is 'x'
if(curr == 'x'):
print(curr, end = "")
return
# Driver code
if __name__ == '__main__':
s = "geekxsforgexxeksxx"
l = len(s)
moveAtEnd(s, 0, l)
# This code is contributed by Shivam Singh
C
// C# implementation to Move all occurrence of letter ‘x’
// from the string s to the end using Recursion
using System;
class GFG{
// Function to move all 'x' in the end
static void moveAtEnd(string s, int i, int l)
{
if (i >= l)
return;
// Store current character
char curr = s[i];
// Check if current character is not 'x'
if (curr != 'x')
Console.Write(curr);
// recursive function call
moveAtEnd(s, i + 1, l);
// Check if current character is 'x'
if (curr == 'x')
Console.Write(curr);
return;
}
// Driver code
public static void Main()
{
string s = "geekxsforgexxeksxx";
int l = s.Length;
moveAtEnd(s, 0, l);
}
}
// This code is contributed by Nidhi_Biet
java 描述语言
<script>
// Javascript implementation to Move
// all occurrence of letter ‘x’ from
// the string s to the end using Recursion
// Function to move all 'x' in the end
function moveAtEnd(s, i, l)
{
if (i >= l)
return;
// Store current character
let curr = s[i];
// Check if current character is not 'x'
if (curr != 'x')
document.write(curr);
// Recursive function call
moveAtEnd(s, i + 1, l);
// Check if current character is 'x'
if (curr == 'x')
document.write(curr);
return;
}
// Driver code
let s = "geekxsforgexxeksxx";
let l = s.length;
moveAtEnd(s, 0, l);
// This code is contributed by suresh07
</script>
Output
geeksforgeeksxxxxx
涉及字符交换的另一个实现:
在这种方法中，我们将交换相邻的字符，以在末尾添加“x”。
下面是上述技术的实现:
C++
// C++ program for above approach
#include<bits/stdc++.h>
using namespace std;
// Recursive program to bring 'x'
// to the end
void rec(char *a, int i)
{
// When the string is completed
// from reverse direction end of recursion
if(i == 0)
{
cout << a << endl;
return;
}
// If the character x is found
if(a[i] == 'x')
{
// Transverse the whole string
int j = i;
while(a[j] != '\0' && a[j+1] != '\0')
{
// Swap the x so that
// it moves to the last
swap(a[j], a[j+1]);
j++;
}
}
// call to the smaller problem now
rec(a, i - 1);
}
// Driver Code
int main()
{
char a[] = {'g', 'e', 'e', 'k', 'x',
's', 'x', 'x', 'k', 's', '\0'};
// Size of a
int n = 10;
// Call to rec
rec(a,n-1);
}
/* This code is contributed by Harsh kedia */
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the
// above approach
import java.util.*;
class Main{
// Recursive program to
// bring 'x' to the end
public static void rec(char a[],
int i)
{
// When the string is completed
// from reverse direction end
// of recursion
if(i == 0)
{
System.out.println(a);
return;
}
// If the character x is found
if(a[i] == 'x')
{
// Transverse the whole string
int j = i;
while(a[j] != '\0' &&
a[j + 1] != '\0')
{
// Swap the x so that
// it moves to the last
char temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
j++;
}
}
// call to the smaller
// problem now
rec(a, i - 1);
}
// Driver code
public static void main(String[] args)
{
char a[] = {'g', 'e', 'e', 'k',
'x', 's', 'x', 'x',
'k', 's', '\0'};
// Size of a
int n = 10;
// Call to rec
rec(a,n-1);
}
}
// This code is contributed by divyeshrabadiya07
Python 3
# Python3 program for above approach
# Recursive program to bring 'x'
# to the end
def rec(a, i):
# When the string is completed
# from reverse direction end
# of recursion
if (i == 0):
a.pop()
print("".join(a))
return
# If the character x is found
if (a[i] == 'x'):
# Transverse the whole string
j = i
while(a[j] != '\0' and
a[j + 1] != '\0'):
# Swap the x so that
# it moves to the last
(a[j], a[j + 1]) = (a[j + 1], a[j])
j += 1
# Call to the smaller problem now
rec(a, i - 1)
# Driver code
if __name__=="__main__":
a = [ 'g', 'e', 'e', 'k', 'x',
's', 'x', 'x', 'k', 's', '\0' ]
# Size of a
n = 10
# Call to rec
rec(a, n - 1)
# This code is contributed by rutvik_56
C
// C# program for the
// above approach
using System;
class GFG
{
// Recursive program to
// bring 'x' to the end
static void rec(char[] a, int i)
{
// When the string is completed
// from reverse direction end
// of recursion
if(i == 0)
{
Console.WriteLine(a);
return;
}
// If the character x is found
if(a[i] == 'x')
{
// Transverse the whole string
int j = i;
while(a[j] != '\0' &&
a[j + 1] != '\0')
{
// Swap the x so that
// it moves to the last
char temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
j++;
}
}
// call to the smaller
// problem now
rec(a, i - 1);
}
// Driver code
static void Main()
{
char[] a = {'g', 'e', 'e', 'k',
'x', 's', 'x', 'x',
'k', 's', '\0'};
// Size of a
int n = 10;
// Call to rec
rec(a,n-1);
}
}
// This code is contributed by divyesh072019
java 描述语言
<script>
// Javascript program for the above approach
// Recursive program to
// bring 'x' to the end
function rec(a, i)
{
// When the string is completed
// from reverse direction end
// of recursion
if(i == 0)
{
document.write(a.join(""));
return;
}
// If the character x is found
if(a[i] == 'x')
{
// Transverse the whole string
let j = i;
while(a[j] != '\0' &&
a[j + 1] != '\0')
{
// Swap the x so that
// it moves to the last
let temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
j++;
}
}
// call to the smaller
// problem now
rec(a, i - 1);
}
let a = ['g', 'e', 'e', 'k', 'x', 's', 'x', 'x', 'k', 's', '\0'];
// Size of a
let n = 10;
// Call to rec
rec(a, n - 1);
// This code is contributed by decode2207.
</script>
Output
geeksksxxx
时间复杂度: O(N)，其中 N 为给定字符串的长度。 辅助空间: O(N)。
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