Dada la string str que consta de varias palabras, la tarea es invertir la string completa palabra por palabra.
Ejemplos:
Entrada: str = «I Love To Code»
Salida: Code To Love I
Entrada: str = «estructuras de datos y algoritmos»
Salida: algoritmos y estructuras de datos
Enfoque: este problema se puede resolver no solo con la ayuda de strtok() , sino que también se puede resolver usando Stack Container Class en STL C++ siguiendo los pasos dados:
- Crea una pila vacía.
- Recorra toda la string, mientras recorre agregue los caracteres de la string en una
variable temporal hasta que obtenga un espacio (‘ ‘) y empuje esa variable temporal a la pila. - Repita el paso anterior hasta el final de la string.
- Saque las palabras de la pila hasta que la pila no esté vacía, lo que será en orden inverso.
A continuación se muestra la implementación del enfoque anterior:
C++
//C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;
//function to reverse the words
//of the given string without using strtok().
void reverse(string s)
{
//create an empty string stack
stack<string> stc;
//create an empty temporary string
string temp="";
//traversing the entire string
for(int i=0;i<s.length();i++)
{
if(s[i]==' ')
{
//push the temporary variable into the stack
stc.push(temp);
//assigning temporary variable as empty
temp="";
}
else
{
temp=temp+s[i];
}
}
//for the last word of the string
stc.push(temp);
while(!stc.empty()) {
// Get the words in reverse order
temp=stc.top();
cout<<temp<<" ";
stc.pop();
}
cout<<endl;
}
//Driver code
int main()
{
string s="I Love To Code";
reverse(s);
return 0;
}
// This code is contributed by Konderu Hrishikesh.
Java
//Java implementation of
// the above approach
import java.util.*;
class GFG{
// Function to reverse the words
// of the given String without
// using strtok().
static void reverse(String s)
{
// Create an empty String stack
Stack<String> stc = new Stack<>();
// Create an empty temporary String
String temp = "";
// Traversing the entire String
for(int i = 0; i < s.length(); i++)
{
if(s.charAt(i) == ' ')
{
// Push the temporary
// variable into the stack
stc.add(temp);
// Assigning temporary
// variable as empty
temp = "";
}
else
{
temp = temp + s.charAt(i);
}
}
// For the last word
// of the String
stc.add(temp);
while(!stc.isEmpty())
{
// Get the words in
// reverse order
temp = stc.peek();
System.out.print(temp + " ");
stc.pop();
}
System.out.println();
}
//Driver code
public static void main(String[] args)
{
String s = "I Love To Code";
reverse(s);
}
}
// This code is contributed by gauravrajput1
Python3
# Python3 implementation of # the above approach # function to reverse the # words of the given string # without using strtok(). def reverse(s): # create an empty string # stack stc = [] # create an empty temporary # string temp = "" # traversing the entire string for i in range(len(s)): if s[i] == ' ': # push the temporary variable # into the stack stc.append(temp) # assigning temporary variable # as empty temp = "" else: temp = temp + s[i] # for the last word of the string stc.append(temp) while len(stc) != 0: # Get the words in reverse order temp = stc[len(stc) - 1] print(temp, end = " ") stc.pop() print() # Driver code s = "I Love To Code" reverse(s) # This code is contributed by divyeshrabadiya07
C#
// C# implementation of
// the above approach
using System;
using System.Collections;
class GFG
{
// Function to reverse the words
// of the given String without
// using strtok().
static void reverse(string s)
{
// Create an empty String stack
Stack stc = new Stack();
// Create an empty temporary String
string temp = "";
// Traversing the entire String
for(int i = 0; i < s.Length; i++)
{
if(s[i] == ' ')
{
// Push the temporary
// variable into the stack
stc.Push(temp);
// Assigning temporary
// variable as empty
temp = "";
}
else
{
temp = temp + s[i];
}
}
// For the last word
// of the String
stc.Push(temp);
while(stc.Count != 0)
{
// Get the words in
// reverse order
temp = (string)stc.Peek();
Console.Write(temp + " ");
stc.Pop();
}
Console.WriteLine();
}
// Driver code
static void Main()
{
string s = "I Love To Code";
reverse(s);
}
}
// This code is contributed by divyesh072019
Javascript
<script>
// Javascript implementation of
// the above approach
// Function to reverse the words
// of the given String without
// using strtok().
function reverse(s)
{
// Create an empty String stack
let stc = [];
// Create an empty temporary String
let temp = "";
// Traversing the entire String
for(let i = 0; i < s.length; i++)
{
if(s[i] == ' ')
{
// Push the temporary
// variable into the stack
stc.push(temp);
// Assigning temporary
// variable as empty
temp = "";
}
else
{
temp = temp + s[i];
}
}
// For the last word
// of the String
stc.push(temp);
while(stc.length != 0)
{
// Get the words in
// reverse order
temp = stc[stc.length - 1];
document.write(temp + " ");
stc.pop();
}
}
let s = "I Love To Code";
reverse(s);
</script>
Producción:
Code To Love I
Complejidad de tiempo: O(N)
Otro enfoque: aquí se analiza un enfoque sin usar la pila . Este problema también se puede resolver usando la pila siguiendo los pasos a continuación:
- Crea una pila vacía.
- Tokenize la string de entrada en palabras usando espacios como separador con la ayuda de strtok()
- Empuje las palabras en la pila.
- Saque las palabras de la pila hasta que la pila no esté vacía, lo que será en orden inverso.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to reverse the words
// of the given sentence
void reverse(char k[])
{
// Create an empty character array stack
stack<char*> s;
char* token = strtok(k, " ");
// Push words into the stack
while (token != NULL) {
s.push(token);
token = strtok(NULL, " ");
}
while (!s.empty()) {
// Get the words in reverse order
cout << s.top() << " ";
s.pop();
}
}
// Driver code
int main()
{
char k[] = "geeks for geeks";
reverse(k);
return 0;
}
Java
// Java implementation of the approach
import java.util.Arrays;
import java.util.Stack;
class GFG {
// Function to reverse the words
// of the given sentence
static void reverse(String k)
{
// Create an empty character array stack
Stack<String> s = new Stack<>();
String[] token = k.split(" ");
// Push words into the stack
for (int i = 0; i < token.length; i++) {
s.push(token[i]);
}
while (!s.empty()) {
// Get the words in reverse order
System.out.print(s.peek() + " ");
s.pop();
}
}
// Driver code
public static void main(String[] args)
{
String k = "geeks for geeks";
reverse(k);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach # Function to reverse the words # of the given sentence def reverse(k): # Create an empty character array stack s = [] token = k.split() # Push words into the stack for word in token : s.append(word); while (len(s)) : # Get the words in reverse order print(s.pop(), end = " "); # Driver code if __name__ == "__main__" : k = "geeks for geeks"; reverse(k); # This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
// Function to reverse the words
// of the given sentence
static void reverse(String k)
{
// Create an empty character array stack
Stack<String> s = new Stack<String>();
String[] token = k.Split(' ');
// Push words into the stack
for (int i = 0; i < token.Length; i++) {
s.Push(token[i]);
}
while (s.Count != 0) {
// Get the words in reverse order
Console.Write(s.Peek() + " ");
s.Pop();
}
}
// Driver code
public static void Main(String[] args)
{
String k = "geeks for geeks";
reverse(k);
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// JavaScript implementation of the approach
// Function to reverse the words
// of the given sentence
function reverse(k)
{
// Create an empty character array stack
let s = [];
let token = k.split(" ");
// Push words into the stack
for (let i = 0; i < token.length; i++) {
s.push(token[i]);
}
while (s.length > 0) {
// Get the words in reverse order
document.write(s[s.length - 1] + " ");
s.pop();
}
}
// Driver code
let k = "geeks for geeks";
reverse(k);
</script>
Producción:
geeks for geeks
Complejidad de tiempo: O(N)
Publicación traducida automáticamente
Artículo escrito por SnehaPasupula y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA