使用堆栈反转字符串的单词

给定由多个单词组成的字符串str ，任务是逐字反转整个字符串。
例子：

Input: str = “I Love To Code” 
Output: Code To Love I
Input: str = “data structures and algorithms” 
Output: algorithms and structures data 
 

方法：这个问题不仅可以在 strtok() 的帮助下解决，还可以通过使用 STL C++ 中的堆栈容器类来解决，具体步骤如下：

1. 创建一个空堆栈。
2. 遍历整个字符串，同时将字符串的字符添加到临时
   变量，直到你得到一个空格（''）并将该临时变量推入堆栈。
3. 重复上述步骤，直到字符串结束。
4. 从堆栈中弹出单词，直到堆栈不为空，这将是相反的顺序。

下面是上述方法的实现：

C++

//C++ implementation of the above approach
#include
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 stc;
 
  //create an empty temporary string
  string temp="";
 
  //traversing the entire string
  for(int i=0;i

---

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 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

 
    // 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);
     

输出：

Code To Love I

时间复杂度： O(N)

另一种方法：这里讨论了一种不使用堆栈的方法。这个问题也可以通过以下步骤使用堆栈来解决：


1. 创建一个空堆栈。
2. 在 strtok() 的帮助下，使用空格作为分隔符将输入字符串标记为单词
3. 将单词压入堆栈。
4. 从堆栈中弹出单词，直到堆栈不为空，这将是相反的顺序。

下面是上述方法的实现：


C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to reverse the words
// of the given sentence
void reverse(char k[])
{
 
    // Create an empty character array stack
    stack 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 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 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.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

 
    // 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);
 

输出：

geeks for geeks

时间复杂度： O(N)