Dada una string str , la tarea es invertir todas las palabras individuales.
Ejemplos:
Entrada: str = “Hola mundo”
Salida: olleH dlroWEntrada: str = «Geeks para Geeks»
Salida: skeeG rof skeeG
Enfoque: En esta publicación se ha discutido una solución al problema anterior . Tiene una complejidad temporal de O(n) y usa O(n) espacio adicional. En esta publicación, discutiremos una solución que usa O(1) espacio adicional.
- Atraviesa la cuerda hasta que nos encontremos con un espacio.
- Después de encontrar el espacio, usamos dos variables ‘inicio’ y ‘fin’ que apuntan al primer y último carácter de la palabra e invertimos esa palabra en particular.
- Repita los pasos anteriores hasta la última palabra.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the string after
// reversing the individual words
string reverseWords(string str)
{
// Pointer to the first character
// of the first word
int start = 0;
for (int i = 0; i <= str.size(); i++) {
// If the current word has ended
if (str[i] == ' ' || i == str.size()) {
// Pointer to the last character
// of the current word
int end = i - 1;
// Reverse the current word
while (start < end) {
swap(str[start], str[end]);
start++;
end--;
}
// Pointer to the first character
// of the next word
start = i + 1;
}
}
return str;
}
// Driver code
int main()
{
string str = "Geeks for Geeks";
cout << reverseWords(str);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static String swap(String str, int i, int j)
{
StringBuilder sb = new StringBuilder(str);
sb.setCharAt(i, str.charAt(j));
sb.setCharAt(j, str.charAt(i));
return sb.toString();
}
// Function to return the string after
// reversing the individual words
static String reverseWords(String str)
{
// Pointer to the first character
// of the first word
int start = 0;
for (int i = 0; i < str.length(); i++)
{
// If the current word has ended
if (str.charAt(i) == ' ' ||
i == str.length()-1 )
{
// Pointer to the last character
// of the current word
int end;
if (i == str.length()-1)
end = i ;
else
end = i - 1 ;
// Reverse the current word
while (start < end)
{
str = swap(str,start,end) ;
start++;
end--;
}
// Pointer to the first character
// of the next word
start = i + 1;
}
}
return str ;
}
// Driver code
public static void main(String args[])
{
String str = "Geeks for Geeks";
System.out.println(reverseWords(str));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the approach # Function to return the after # reversing the individual words def reverseWords(Str): # Pointer to the first character # of the first word start = 0 for i in range(len(Str)): # If the current word has ended if (Str[i] == ' ' or i == len(Str)-1): # Pointer to the last character # of the current word end = i - 1 if(i == len(Str)-1): end = i # Reverse the current word while (start < end): Str[start], Str[end]=Str[end],Str[start] start+=1 end-=1 # Pointer to the first character # of the next word start = i + 1 return "".join(Str) # Driver code Str = "Geeks for Geeks" Str=[i for i in Str] print(reverseWords(Str)) # This code is contributed by mohit kumar 29
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the string after
// reversing the individual words
// Function to return the string after
// reversing the individual words
static String reverseWords(String str)
{
// Pointer to the first character
// of the first word
int start = 0;
for (int i = 0; i < str.Length; i++)
{
// If the current word has ended
if (str[i] == ' ' ||
i == str.Length-1 )
{
// Pointer to the last character
// of the current word
int end;
if (i == str.Length-1)
end = i ;
else
end = i - 1 ;
// Reverse the current word
while (start < end)
{
str = swap(str,start,end) ;
start++;
end--;
}
// Pointer to the first character
// of the next word
start = i + 1;
}
}
return str ;
}
static String swap(String str, int i, int j)
{
char []ch = str.ToCharArray();
char temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
return String.Join("",ch);
}
// Driver code
public static void Main(String []args)
{
String str = "Geeks for Geeks";
Console.WriteLine(reverseWords(str));
}
}
/* This code is contributed by PrinciRaj1992 */
Javascript
<script>
// Javascript implementation of the approach
function swap(str, i, j)
{
let sb = str.split("");
sb[i] = str[j];
sb[j] = str[i];
return sb.join("");
}
// Function to return the string after
// reversing the individual words
function reverseWords(str)
{
// Pointer to the first character
// of the first word
let start = 0;
for(let i = 0; i < str.length; i++)
{
// If the current word has ended
if (str[i] == ' ' ||
i == str.length-1 )
{
// Pointer to the last character
// of the current word
let end;
if (i == str.length-1)
end = i ;
else
end = i - 1 ;
// Reverse the current word
while (start < end)
{
str = swap(str, start, end);
start++;
end--;
}
// Pointer to the first character
// of the next word
start = i + 1;
}
}
return str;
}
// Driver code
let str = "Geeks for Geeks";
document.write(reverseWords(str));
// This code is contributed by unknown2108
</script>
Producción:
skeeG rof skeeG
Complejidad temporal: O(n)
Espacio auxiliar: O(1)